M.C.Q (1 Marks) - Physics STD 12 Science Questions - Vidyadip

Questions

M.C.Q (1 Marks)

🎯

Test yourself on this topic

20 questions · timed · auto-graded

MCQ 11 Mark

The work function of a metal is $hv_{0.}$ Light of frequency $v$ falls on this metal. The photoelectric effect will take place only if$:$

✓
$\text{v}\geq\text{v}_0$
B
$\text{v}>2\text{v}_0$
C
$\text{v}<\text{v}_0$
D
$\text{v}<\frac{\text{v}_0}{2}$

Answer

Correct option: A.

$\text{v}\geq\text{v}_0$

As the work function of the metal is $hv_0,$ the threshold frequency of the metal is $v_0.$
For photoelectric effect to occur, the frequency of the incident light should be greater than or equal to the threshlod frequency of the metal on which light is incident.

View full question & answer→

Question 21 Mark

Light of wavelength $\lambda$ falls on a metal having work function $\frac{\text{h}_\text{c}}{\lambda_0}.$ Photoelectric effect will take place only if:

$\lambda\geq\lambda_0$
$\lambda\geq2\lambda_0$
$\lambda\leq\lambda_0$
$\lambda<\frac{\lambda_ 0}{2}.$

Answer

$\lambda\leq\lambda_0$

Explanation:
As the work-function of the metal is $\frac{\text{hc}}{\lambda_0}$, its threshold wavelength is $\lambda_0.$
For photoelectric effect, the wavelength of the incident light should be less than or equal to the threshold wavelength of the metal on which light is incident.

View full question & answer→

MCQ 31 Mark

A proton and an electron are accelerated by the same potential difference. Let $\lambda_\text{e}$ and $\lambda_\text{p}$ denote the de Broglie wavelengths of the electron and the proton respectively.

A
$\lambda_\text{e}=\lambda_\text{p}.$
B
$\lambda_\text{e}<\lambda_\text{p}.$
✓
$\lambda _\text{e}>\lambda_\text{p}.$
D
The relation between $\lambda_\text{e}$, and $\lambda_\text{p}$ depends on the accelerating potential difference.

Answer

Correct option: C.

$\lambda _\text{e}>\lambda_\text{p}.$

Let $m_e$ and $m_p$ be the masses of electron and proton, respectively.
Let the applied potential difference be $V.$
Thus, the de$-$Broglie wavelength of the electron,
$\lambda=\frac{\text{h}}{\sqrt{2\text{m}_\text{e}\text{eV}}}\dots(1)$
And de$-$Broglie wavelength of the proton,
$\lambda=\frac{\text{h}}{\sqrt{2\text{m}_\text{p}\text{eV}}}\dots(2)$
Dividing equation $(2)$ by equation $(1),$ we get$:$
$\frac{\lambda_\text{p}}{\lambda_\text{e}}=\frac{\sqrt{\text{m}_\text{e}}}{\sqrt{\text{m}_\text{p}}}$
$\text{m}_\text{e}<\text{m}_\text{p}$
$\therefore\frac{\lambda_\text{p}}{\lambda_\text{e}}<1$
$\Rightarrow\lambda_ \text{p}<\lambda_\text{e}$

View full question & answer→

Question 41 Mark

Photoelectric effect supports quantum nature of light because:

There is a minimum frequency below which no photoelectrons are emitted.
The maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity.
Even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately.
Electric charge of the photoelectrons is quantized.

Answer

There is a minimum frequency below which no photoelectrons are emitted.
The maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity.
Even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately:

Explanation:
Photoelectric effect can be explained on the basis of quantum nature of light. According to the quantum nature of light, energy in light is not uniformly spread. It is contained in packets or quanta known as photons.
Energy of a photon, E = hv, where h is Planck's constant and v is the frequency of light.
Above a particular frequency, called threshold frequency, energy of a photon is sufficient to emit an electron from the metal surface and below which, no photoelectron is emitted, as the energy of the photon is low. Hence, option (a) supports the quantum nature of light.
Now, kinetic energy of an electron,
$\text{K}=\text{hv}_0-\varphi$
Thus, kinetic energy of a photoelectron depends only on the frequency of light (or energy). This shows that if the intensity of light is increased, it only increases the number of photons and not the energy of photons. Kinetic energy of photons can be increased by increasing the frequency of light or by increasing the energy of photon, which supports E = hv and, hence, the quantum nature of light. Hence, option (b) also supports the quantum nature of light.
hotoelectrons are emitted from a metal surface even if the metal surface is faintly illuminated; it means that less photons will interact with the electrons. However, few electrons absorb energy from the incident photons and come out from the metal. This shows the quantum nature of light. Hence, (c) also supports the quantum nature of light.
Electric charge of the photoelectrons is quantised; but this statement does not support the quantum nature of light.

View full question & answer→

Question 51 Mark

Planck constant has the same dimensions as:

Force × time.
Force × distance.
Force × speed.
Force × distance × time.

Answer

Force × distance × time.

Explanation:
Planck's constant
$\text{h}=\frac{\text{E}}{\text{v}}=\frac{\text{Force}\times\text{distanace}}{\text{frequency}}$
$\Rightarrow\text{h}=\text{force}\times\text{distance}\times{\text{times}}.$

View full question & answer→

MCQ 61 Mark

If the wavelength of light in an experiment on photoelectric effect is doubled$:$

A
Photoelectric emission will not take place.
B
Photoelectric emission may or may not take place.
C
The stopping potential will increase.
✓
The stopping potential will decrease.

Answer

Correct option: D.

The stopping potential will decrease.

For photoelectric effect to be observed, wavelength of incident light should not be more than the largest wavelength called threshold wavelength $(\lambda_0)$ If the wavelength of light in an experiment on photoelectric effect is doubled and if it is equal to or less than the threshold wavelength, then photoelectric emission will take place. If it is greater than the threshold wavelength, photoelectric emission will not take place. The photoelectric emission may or may not take place.Photoelectric emission depends on the wavelength of incident light.
Hence, option $(b)$ is correct and $(a)$ is incorrect.
From Einstein's photoelectric equation,
$\text{eV}_0=\frac{\text{hc}}{\lambda_0}-\varphi,$
where $V_0 =$ stopping potential
$\lambda_0 =$ threshold wavelength
$h =$ Planck's constant
$\varphi =$ work$-$function of metal It is clear that
$\text{V}_0\propto\frac{1}{\lambda_0}$
Thus, if the wavelength of light in an experiment on photoelectric effect is doubled, its stopping potential will become half.

View full question & answer→

Question 71 Mark

The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photocurrent is recorded. An electric field is switched on which has a vertically downward direction.

The photocurrent will increase.
The kinetic energy of the electrons will increase.
The stopping potential will decrease.
The threshold wavelength will increase.

Answer

The kinetic energy of the electrons will increase.

Explanation:
As there is no effect of electric field on the number of photons emitted, the photoelectric current will remain same. Hence, option (a) is incorrect.
When an electric field is applied, then electric force will act on the electron moving opposite the direction of electric field, which will increase the kinetic energy of the electron. Hence, option (b) is correct.
As the kinetic energy of the electron is increasing, its stopping potential will increase. Hence, option (c) is incorrect.
Threshold wavelength is the characteristic property of the metal and will not change. Hence, (d) is incorrect.

View full question & answer→

Question 81 Mark

The photocurrent in an experiment on photoelectric effect increases if:

The intensity of the source is increased.
The exposure time is increased.
The intensity of the source is decreased.
The exposure time is decreased.

Answer

The intensity of the source is increased.

Explanation:
When the intensity of the source is increased, the number of photons emitted from the source increases. As a result, a large number of electrons of the metal interact with these photons and hence, the number of electrons emitted from the metal increases. Thus, the photocurrent in an experiment of photoelectric effect increases. The photocurrent does not depend on the exposure time.

View full question & answer→

Question 91 Mark

When the intensity of a light source is increased:

The number of photons emitted by the source in unit time increases.
The total energy of the photons emitted per unit time increases.
More energetic photons are emitted.
Faster photons are emitted.

Answer

The number of photons emitted by the source in unit time increases.
The total energy of the photons emitted per unit time increases.

Explanation:
When the intensity of a light source in increased, a large number of photons are emitted from the light source. Hence, option (a) is correct.
Due to increase in the number of photons, total energy of the photons emitted per unit time also increases. Hence, option (b) is also correct.
Increase in the intensity of light increases only the number of photons, not the energy of photons, Hence, option (c) is incorrect.
The speed of photons is not affected by the intensity of light, Hence, option (d) is incorrect.

View full question & answer→

MCQ 101 Mark

A non.monochromatic light is used in an experiment on photoelectric effect. The stopping potential$:$

A
Is related to the mean wavelength.
B
Is related to the longest wavelength.
✓
Is related to the shortest wavelength.
D
Is not related to the wavelength.

Answer

Correct option: C.

Is related to the shortest wavelength.

For photoelectric effect to be observed, wavelength of the incident light $(\lambda)$ should be less than the threshold wavelength $(\lambda _0)$ of the metal.
Einstein's photoelectric equation:
$\text{eV}_0=\frac{\text{hc}}{\lambda}-\varphi$
Here, $V_0 =$ stopping potential.
$\lambda_0 =$ threshold wavelength.
$h =$ Planck's constant.
$\varphi =$ work$-$function of metal.

View full question & answer→

Question 111 Mark

Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased:

Both p and E increase.
P increases and E decreases.
P decreases and E increases.
Both p and E decrease.

Answer

Both p and E increase.

Explanation:
From the de-Broglie relation, wavelength,
$\lambda=\frac{\text{h}}{\text{p}}\dots(1)$
$\Rightarrow\text{p}=\frac{\text{h}}{\lambda}$
Here, h = Planck's constant
p = momentum of electron
It is clear from the above equation that $\text{p}\alpha\frac{1}{\lambda}.$
Thus, if the wavelength $\lambda$ is decreased, then momentum (p) will be increase.
Relation between momentum and energy:
$\text{p}=\sqrt{2\text{mE}}$
Here, E = energy of electron
m = mass of electron
Substituting the value of p in equation (1), we get:
$\lambda=\frac{\text{h}}{\sqrt{2\text{mE}}}$
$\Rightarrow\sqrt{\text{E}}=\frac{\text{h}}{\lambda\sqrt{2\text{m}}}$
$\Rightarrow\text{E}=\frac{\text{h}^2}{2\text{m}\lambda^2}$
Thus, on decreasing $\lambda,$ the energy will increase.

View full question & answer→

MCQ 121 Mark

If the frequency of light in a photoelectric experiment is doubled, the stopping potential will:

A
Be doubled.
B
Be halved.
✓
Become more than double.
D
Become less than double.

Answer

Correct option: C.

Become more than double.

According to Einstein's equation of photoelectric effect,
$\text{eV}_0=\text{hv} - \varphi$
$\Rightarrow\text{V}_0=\frac{\text{hv}- \varphi}{\text{e}}\dots(1)$
Here, $V_0 =$ stopping potential
$v =$ frequency of light
$\varphi =$ work function
Let the new frequency of light be $2ν$ and the corresponding stopping potential be $V_0\ '.$
Therefore,
$\text{eV}_0\ '=2\text{hv}-\varphi$
$\text{V}_0\ '=\frac{2\text{hv}-\varphi}{\text{e}}\dots(2)$
Multiplying both sides of equation $(1)$ by $2,$ we get:
$2\text{V}_0=\frac{2\text{hv}-2\varphi}{\text{e}}\dots(3)$
Now if we compare $(2)$ and $(3),$ it can be observed that:
$\frac{2\text{hv}-\varphi}{\text{e}}>\frac{2\text{hv}-2\varphi}{\text{e}}$
$\Rightarrow\text{V}_0\ '>2\text{V}_0$
It is clear from the above equation that if the frequency of light in a photoelectric experiment is doubled,
the stopping potential will be more than doubled.

View full question & answer→

MCQ 131 Mark

In which of the following situations, the heavier of the two particles has smaller de Broglie wavelength? The two particles:

✓
Move with the same speed.
B
Move with the same linear momentum.
C
Move with the same kinetic energy.
D
Have fallen through the same height.

Answer

Correct option: A.

Move with the same speed.

Let $m_1$ be the mass of the heavier particle and $m_2$ be the mass of the lighter particle.
If both the particles are moving with the same speed $v,$ de Broglie wavelength of the heavier particle,
$\lambda_1=\frac{\text{h}}{\text{m}_1\text{v}}\dots(1)$
de Broglie wavelength of the lighter particle,
$\lambda=\frac{\text{h}}{\text{m}_2\text{v}}\dots(2)$
Thus, from equations $(1)$ and $(2),$
we find that if the particles are moving with the same speed $v, $
then $\lambda _1<\lambda_2.$

View full question & answer→

Question 141 Mark

A photon of energy hv is absorbed by o free electron of a metal having work function $\varphi<\text{hv}.$

The electron is sure to come out.
The electron is sure to come out with a kinetic energy $\text{hv}-\varphi.$
Either the electron does not come out or it comes out with a kinetic energy $\text{hv}-\varphi$ .
It may come out with a kinetic energy less than $\text{hv}-\varphi.$

Answer

It may come out with kinetic energy less than $\text{hv}-\varphi.$

Explanation:
When light is incident on the metal surface, the photons of light collide with the free electrons. In some cases, a photon can give all the energy to the free electron. If this energy is more then the work-function of the metal,then there are two possibilities. The electron can come out of the metal with kinetic energy $\text{hv}-\varphi.$ or it may lose energy on collision with the atoms of the metal and come out with kinetic energy less than $\text{hv}-\varphi.$ Thus, it may come out with kinetic energy less than $\text{hv}-\varphi.$

View full question & answer→

MCQ 151 Mark

Let $n_r$ and $n_b$ be the number of photons emitted by a red bulb and a blue bulb, respectively, of equal power in a given time$:$

A
$n_r = n_{b}.$
B
$n_r< n_{b}.$
✓
$n_r > n_{b}.$
D
The information is insufficient to derive a relation between $n_r$ and $n_{b}.$

Answer

Correct option: C.

$n_r > n_{b}.$

The two bulbs are of equal power. It means that they consume equal amount of energy per unit time.
Now, as the frequency of blue light $(fb)$ is higher than the frequency of red light $(f_r).$ $hf_b > hf_r.$
Hence, the energy of a photon of blue light is more than the energy of a photon of red light.
Thus, a photon of blue light requires more energy than a photon of red light to be emitted.
For the same energy given to the bulbs in a certain time, the number of photons of blue light will be less than that of red light.
$\therefore\text{n}_\text{r}>\text{n}_\text{b} ($As the amount of energy emitted from the two bulb is same$)$

View full question & answer→

Question 161 Mark

When stopping potential is applied in an experiment on photoelectric effect, no photocurrent is observed. This means that:

The emission of photoelectrons is stopped.
The photoelectrons are emitted but are reabsorbed by the emitter metal.
The photoelectrons are accumulated near the collector plate.
The photoelectrons are dispersed from the sides of the apparatus.

Answer

The photoelectrons are emitted but are re-absorbed by the emitter metal.

Explanation:
In an experiment on photoelectric effect, the photons incident at the metal plate cause photoelectrons to be emitted. The metal plate is termed as "emitter". The electrons ejected are collected at the other metal plate called "collector". When the potential of the collector is made negative with respect to the emitter (or the stopping potential is applied), the electrons emitted from the emitter are repelled by the collector.
As a result, some electrons go back to the cathode and the current decreases.

View full question & answer→

MCQ 171 Mark

The frequency and intensity of a light source are both doubled. Consider the following statements.

The saturation photocurrent remains almost the same.
The maximum kinetic energy of the photoelectrons is doubled.

A
Both $A$ and $B$ are true.
✓
$A$ is true but $B$ is false.
C
$A$ is false but $B$ is true.
D
Both $A$ and $B$ are false.

Answer

Correct option: B.

$A$ is true but $B$ is false.

Saturated current varies directly with the intensity of light. As the intensity of light is increased, a large number of photons fall on the metal surface. As a result, a large number of electrons interact with the photons. As a result, the number of emitted electrons increases and, hence, the current also increases.
At the same time, the frequency of the light source also increases.Also, with the increase in frequency of light, the stopping potential increases as well. This will reduce the current. The combined effect of these two is that the current will remain the same.
From the Einstein's photoelectric equation.
$\text{K}_\text{max}= \text{hv}-\varphi$
Where $K_{max} =$ kinetic energy of electron
$v =$ frequency of light
$\varphi =$ work function of metal
It is clear from the above equation.
As the frequency of light source is doubled, kinetic energy of electron increases.
But, it becomes more than the double.

View full question & answer→

Question 181 Mark

Two photons having:

Equal wavelengths have equal linear momenta.
Equal energies have equal linear momenta.
Equal frequencies have equal linear momenta.
Equal linear momenta have equal wavelengths.

Answer

Equal linear momenta have equal wavelengths.

Explanation:
Two photons having equal linear momenta have equal wavelengths is correct. As in the rest of the options magnitude of momentum or energy can be same because energy and momentum are inversely proportional to wavelength. But the direction of propagation of the photons can be different.

View full question & answer→

Question 191 Mark

A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential:

Will increase.
Will decrease.
Will remain constant.
Will either increase or decrease.

Answer

Will remain constant.

Explanation:
As the source is removed farther from the emitting metal, the intensity of light will decrease. As the stopping potential does not depend on the intensity of light, it will remain constant.

View full question & answer→

Question 201 Mark

The equation E = pc is valid:

For an electron as well as for a photon.
For an electron but not for a photon.
For a photon but not for an electron.
Neither for an electron nor for a photon.

Answer

For a photon but not for an electron.

Explanation:
The equation E = pc is valid for a particle with zero rest mass. The rest mass of a photon is zero, but the rest mass of an electron is not zero. So, the equation will be valid for photon, and not electron.

View full question & answer→

Generate Paper Free All Photoelectric Effect and WaveParticle Duality topics