Physics 4 Marks Questions

Alternate Answer

1. Nuclear fission of E to D and C; as there is a increase in binding energy per nucleon.
2. Nuclear fusion of A and B into C; as there is a increase in binding energy per nucleon.

$\text{First step} - \propto\text{ particle Second step} – \beta\text{ particle}$.

1. (d)

Explanation:

Rutherford's atom had a positively charged centre and electrons were revolving outside it. It is also called the planetary model of the atom, as in option (d).

2. (d) Most $\alpha$-particle will not suffer more than 1º scattering during passage through gold foil.

Explanation:

As the gold foil is very thin, it can be assumed that $\alpha$-particles will suffer not more than one scattering during their passage through it. Therefore, computation of the trajectory of an $\alpha$-particle scattered by a single nucleus is enough.

3. (c) Impact parameter

Explanation:

Trajectory of $\alpha$-particles depends on impact parameter, which is the perpendicular distance of the initial velocity vector of the $\alpha$particles from the centre of the nucleus. For small impact parameter, $\alpha$ particle close to the nucleus suffers larger scattering.

4. (b) Minimum

Explanation:

At minimum impact parameter, $\alpha$ particles rebound back$(\theta\approx\pi)$ and suffers large scattering.

5. (d) Both (a) and (b).

Explanation:

In case of head-on-collision, the impact parameter is minimum and the $\alpha$-particle rebounds back. So, the fact that only a small fraction of the number of incident particles rebound back indicates that the number of $\alpha$-particles undergoing head-on collision is small. This in turn implies that the mass of the atom is concentrated in a small volume. Hence, option (a) and (b) are correct.

1. (c) 4

Explanation:

$\text{E}_{\text{n}}=\frac{-13.6}{\text{n}^2}(\text{Z}^2)$

In first excited state, EH₂ = 3.4 eV and E_He = -13.6 eV

So, H₂ atom gives excitation energy (13.6 - 3.4 = 10.2 eV) to helium atom

Now, energy of He ion = -13.6 + 10.2 = -3.4 eV

Again, $\text{E}=\frac{-13.6}{\text{n}^2}\times\text{Z}^2$

$\Rightarrow-3.4=\frac{-13.6}{\text{n}^2}\times({2})^2$

$\Rightarrow\text{n}=4$

2. (c) 4.8 × 10^-7 m

Explanation:

$\frac{1}{\lambda}=\frac{13.6\text{Z}^2}{\text{hc}}\Big[\frac{1}{\text{h}^2_1}-\frac{1}{\text{h}^2_2}\Big]$

Here, n₁ = 3 and n₂ = 4

$\Rightarrow\lambda=4.8\times10^{-7}\text{m}$

3. (a) $\frac{1}{4}$

Explanation:

Kinetic energy, $\text{K}\propto\frac{\text{z}^2}{\text{n}^2}$

$\frac{\text{K}_{\text{H}_{2}}}{\text{K}_{\text{H}_{\text{e}}}}=\bigg(\frac{\text{Z}_{\text{H}_{2}}}{\text{Z}_{\text{He}}}\bigg)^2=\big(\frac{1}{2}\big)^2=\frac{1}{4}$

4. (b) $\frac{\text{h}^2\in_0}{\pi\text{me}^2}$

Explanation:

Radius of the permitted orbit is r $=\frac{\text{n}^2\text{h}^2\in_0}{\pi\text{mze}^2}$

For hydrogen atom in ground state, i.e.,

n = 1,Z = 1 $\Rightarrow\text{r}==\frac{\text{n}^2\in_0}{\pi\text{me}^2}$

5. (a) $\frac{\text{h}}{\pi}$

Explanation:

Angular momentum for hydrogen atom is $\text{L}=\frac{\text{nh}}{\pi2}$

For 6 first excited state, n = 2, $\text{L}=\frac{\text{h}}{\pi}$

1. (b) $\frac{4\in_0\text{h}^2}{\text{m}\pi\text{e}^2}$

Explanation:

On other planet : $\text{mvr}=2\text{n}\frac{\text{h}}{2\pi}\Rightarrow\frac{\text{nh}}{\pi\text{mr}}$ 

$\frac{\text{mv}^2}{\text{r}}=\frac{1}{4\pi\in_0}\frac{\text{e}^2}{\text{r}^2}$

$\Rightarrow\frac{\text{mn}^2\text{h}^2}{\text{n}^2\text{m}^2\text{r}^3}=\frac{1}{4\pi\in_0}\frac{\text{e}^2}{\text{r}^2}$

Putting n = 1, we get $\text{r}=\frac{4\text{h}^2\in_0}{\text{m}\pi\text{e}^2}$

2. (b) $\frac{\text{v}_0}{2}$

Explanation:

On our planet:$\text{v}_0=\frac{\text{e}}{2\in_{\text{nh}}}$

On other planet:$\text{v}=\frac{\text{e}^2}{2\in_0\big(2\text{n}\big)\text{h}}=\frac{\text{v}_0}{2}$

3. (b) 3.4eV

Explanation:

On our planet:$\text{E}_\text{n}=-\frac{13.6}{\text{n}^2}$

On other planet:$\text{E}_\text{n}=-\frac{13.6}{\big(\text{2n}\big)^2}$

$\Rightarrow\text{E}_\text{n}=\frac{\text{E}_\text{n}}{4}=-3.4\text{eV}$

4. (c) Only (ii) and (iii) are correct.

Explanation:

Centripetal acceleration $=\frac{\text{mv}^2}{\text{r}}$

Further, as n increases, r also increases. Therefore, centripetal acceleration for n = 2 is less than that for n = 1. So, statement (i) is wrong. Statement (ii) and (iii) are correct

5. (c) PE increases, TE increases

Explanation:

Potential energy $=\frac{-\text{C}}{\text{r}^2}$ total energy $=\frac{\text{Rhc}}{\text{n}^2}$ With higher orbit, both r and n increase. So, both become less negative hence both increase.

1. (d) $\infty$

Explanation:

Number of spectral lines in hydrogen atom is $\infty$

2. (d) Lyman series

Explanation:

Lyman series lies in the ultraviolet region

3. (c) 4861 A

Explanation:

The shortest Balmer line has energy = 1|(3.4 - 1.51)|1eV = 1.89eV

and the highest energy = 1(0 - 3.4)1 = 3.4eV The corresponding wavelengths are

$\frac{12400\text{eVA}}{1.89\text{eV}}=6561\text{A}\text{ and }\frac{12400\text{eVA}}{3.4\text{eV}}=3647\text{A}$

Only 4861A is between the first and last line of the Balmer series.

4. (a) A universal constant.
5. (c) 6

1. (c) Quantum numbers
2. (d) Inversely proportional to n³

Explanation:

$\omega=\frac{\text{v}}{\text{r}}$ Further $\omega\propto\frac{1}{\text{n}}$ and $\text{r}\propto\text{n}^2$ 

Hence $\omega\propto\Big(\frac{1}{\text{n}^3}\Big)$

3. (c) $\frac{\text{3h}}{2\pi}$
4. (b) The least energy

Explanation:

The energy of n^th Bohr orbit in hydrogen atom is

$\text{E}_\text{n}=-\frac{13.6}{\text{n}^2}\text{eV}$

For lowest orbit, n = 1

$\therefore\text{E}_1=13.6\text{ eV}$

Thus, the lowest Bohr orbit in hydrogen atom has the least energy.

5. (b) The angular momentum of the electron can only be an integral multiple of $\frac{\text{h}}{2\pi}$.

1. (a) 1215.4 A

Explanation:

From, $\text{V}=\frac{1}{\lambda}=\text{R}\Big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$

n₁ = 1, n₂ = 2 for first spectral line of Lyman series,

$\frac{1}{\lambda}=1.097\times10^7\bigg(\frac{1}{1^2}-\frac{1}{1^2}\bigg)$

$=\frac{3\times1.097\times10^7}{4}\text{m}^{-1}$

$\lambda=\frac{4\times10^{-7}\text{m}}{3\times1.097}$

$=\frac{4000}{3\times1.0947}\text{A}=1215.4\text{A}$

2. (d) 911.6 A

Explanation:

For wavelength limit, we put $\text{n}_2=\infty$

$\therefore\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{1^2}-\frac{1}{\infty}\Big)$

$\lambda=\frac{1}{1.097\times10^7}$

$\text{m}=\frac{1000}{1.097}\text{A}=911.6\text{ A}$

3. (b) 4.57 × 10¹⁴ Hz

Explanation:

For first line of Balmer series, n₁ = 2, n₂ = 3

$\text{V}=\frac{1}{\lambda}=\text{R}\Bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Bigg)$

$\text{V}=\frac{\text{c}}{\lambda}=\text{Rc}\Bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Bigg)$

= 1.097 × 10⁷ × 3 × 10⁸$\Bigg(\frac{1}{2_2}-\frac{1}{3_2}\Bigg)$

= 1.097 × 3 × 10¹⁵ × 3 $\frac{5}{36}$ 4.57 × 10¹⁴ Hz

4. (d) n = 2 to n = 1

Explanation:

$\text{h}\upsilon_{2\rightarrow1}=-13.6\Bigg(\frac{1}{2_2}-\frac{1}{1_2}\Bigg)\text{eV}=10.2\text{eV}$ 

Emission is n = 2 → n = 1 i.e., higher n to lower n. Transition from lower to higher levels are absorption lines.

$13.6\Bigg(\frac{1}{6_2}-\frac{1}{2_2}\Bigg)=+13.6\times\frac{2}{9}$

This is < E_n = 2 → E_n = 1

5. (a) 5 : 9

Explanation:

$\frac{1}{\lambda_\text{max}}=\text{R}\bigg[\frac{1}{2^2}-\frac{1}{3^2}\bigg]=\frac{5\text{R}}{36}$

$\frac{1}{\lambda_\text{max}}=\text{R}\bigg[\frac{1}{2^2}-\frac{1}{\infty}\bigg]=\frac{\text{R}}{4}$

$\Rightarrow\frac{\lambda_\text{max}}{\lambda_\text{max}}=\frac{5\text{R}}{36}\times\frac{4}{\text{R}}=\frac{5}{9}$

1. (d) VI

Explanation:

For Balmer series, n₁ = 2; n₂ = 3, 4,... (lower) (higher)

Therefore, in transition (VI), photon of Balmer series is absorbed

2. (c) 487nm

Explanation:

ln transition II,

E₂ = -3.4 eV, E₄ = -0.85 eV

$\triangle\text{E}=2.55\text{ eV}\Rightarrow\triangle\text{E}=\frac{\text{hc}}{\gamma}$

$\Rightarrow\gamma=\frac{\text{hc}}{\triangle \text{E}}=487\text{nm}$

3. (d) V

Explanation:

Wavelength of radiation = 1030 A

$\triangle\text{E}=\frac{12400}{1030\text{ A}}=120\text{ eV}$

So, difference of energy should be 12.0 eV (approx.) Hence for n₁ = 1 to n₂ = 3

En3 -En1 = -1.51eV -(-13.6eV) $\approx$ 12eV

Therefore, transition V will occur.​​​​​​

4. (a) n₁ = 4,n₂ = 2

Explanation:

$\text{T}^2\propto\text{r}^3\text{and}\text{ r}\propto\text{n}^2$

$\Rightarrow\text{T}^2\propto\text{n}^6\Rightarrow\text{T}\propto\text{n}^3$

$\frac{\text{T}_1}{\text{T}_2}=\Big(\frac{\text{n}_1}{\text{n}_2}\Big)^3$

$\Rightarrow8=\Big(\frac{\text{n}_1}{\text{n}_2}\Big)^3\text{or}\frac{\text{n}_1}{\text{n}_2}=3$

5. (b) If we measure the frequencies of light emitted due to transitions between excited states and the first excited state.