Physics M.C.Q (1 Marks)

$\Rightarrow(\text{Z}-\text{b})^2=\frac{\text{v}}{\text{a}^2}$

2. Acelerating voltage.

Explanation:

Cutoff wavelength (λ_min) is given by

$\lambda_{\text{min}}=\frac{\text{hc}}{\text{eV}}$

Here,

h = Planck's constant

c = Speed of light

V = Accelerating voltage

e = Charge of electron

Clearly, a cutoff wavelength depends on accelerating voltage. It does not depend on the target material, separation between the target and the temperature of the filament.

4. Method of creation.

Explanation:

Let $\lambda_1$ and $\lambda_2$ be the wavelengths of the photons of continuous and characteristic X-rays, respectively.

Given:

$\lambda_1=\lambda_2=\lambda$

Now, frequency $(\nu)$ is given by

$\nu=\frac{\text{c}}{\lambda_1}=\frac{\text{c}}{\lambda_2}$

$\Rightarrow\nu=\frac{\text{c}}{\lambda}$

Hence, the frequency of both the photons is the same.

Energy of a photon (E) is given by

$\text{E}=\text{h}\nu$

As the frequency of both the photons is the same, they will have the same energy.

The photon of the continuous X-ray and the photon of the characteristic X-ray consist of the same penetrating power.

The photon of the characteristic X-ray is created because of the transition of an electron from one shell to another. The photon of the continuous X-ray is created because of the conversion of the kinetic energy of an electron into a photon of electromagnetic radiation.

2. V_A > V_B, Z_A < Z_B

Explanation:

It is clear from the figure that the X-ray of tube A has less cutoff wavelength than the X-ray of tube B.

$\therefore\lambda_\text{A}<\lambda_\text{B}$

Using Moseley's Law,

Z_A < Z_B

$\lambda\propto\frac{1}{\text{V}}$, where V is the voltage applied in the X-ray tube.

$\therefore$ V_A > V_B

3. The frequency is higher.
4. The photon energy is higher.

Explanation:

Harder X-rays are the X-rays having low wavelengths. Since the frequency varies inversely with the wavelength, hard X-rays have high frequency.

Energy of a photon (E) is given by

$\text{E}=\frac{\text{hc}}{\lambda}$

Here,

h = Planck's constant

c = Speed of light

$\lambda$ = Wavelength of light.

Clearly, energy varies inversely with wavelength. Therefore, the energy of the photon will be higher for the hard X-ray.

3. $\lambda(\text{K}_\alpha)>\lambda(\text{K}_\beta)>\lambda(\text{K}_\gamma)$

4. $\lambda(\text{M}_\alpha)>\lambda(\text{K}_\beta)>\lambda(\text{K}_\alpha)$

Explanation:

The $\text{K}_\gamma$ transition (from the N shell to the K shell) involves more energy than the $\text{K}_\beta$ transition (from the M shell to the K shell), which has more energy than the $\text{K}_\alpha$ transition (from the L shell to the K shell).

As the energy varies inversely with the wavelength,

$\lambda(\text{K}_\alpha)>\lambda(\text{K}_\beta)>\lambda(\text{K}_\gamma)$

The $\text{M}_\alpha$ transition is due to the jumping of an electron from the N shell to the M shell and involves less energy than the $ \text{L}_\alpha$ transition (from the M shell to the L shell), which involves less energy than the $\text{K}_\alpha$ transition (from the L shell to the K shell).

As the energy varies inversely with the wavelength,

$\lambda(\text{M}_\alpha)>\lambda(\text{L}_\alpha)>\lambda(\text{K}_\alpha)$

3. The intensity remains unchanged.
4. The minimum wavelength decreases.

Explanation:

Cutoff wavelength $(\lambda_\text{min})$ is given by

$\lambda_\text{min}=\frac{\text{hc}}{\text{eV}}$

Here,

h = Planck's constant

c = Speed of light

V = Accelerating voltage

e = Charge of electron

If the potential difference applied to an X-ray tube (V) is increased, then the minimum wavelength (cutoff wavelength) gets decreased.

Intensity is not affected by the potential difference applied.

2. K X-ray is emitted when a hole makes a jump from the K shell to some other shell.
3. The wavelength of K X-ray is smaller than the wavelength of L X-ray of the same material.

Explanation:

Energy of a vacant atom is higher than that of a neutral atom.

Hence, option (a) is incorrect.

K X-ray is emitted when an electron makes a jump to the K shell from some other shell. As a result, a positive charge hole is created in the outer shell. As the electron continuously moves to the K shell, the hole moves from the K shell to some other shell.

Hence, option (b) is correct.

K X-ray is emitted due to the transition of an electron from the L or M shell to the K shell and L X-ray is emitted due to the transition of an electron from the M or N shell to the L shell. The energy involved in the transition from the L or M shell to the K shell is higher than the energy involved in the transition from the M or N shell to the L shell. Since the energy is inversely proportional to the wavelength, the wavelength of the K X-ray is smaller than the wavelength of the L X-ray of the same material. Hence, option (c) is correct.

If E_K, E_L and E_M are the energies of K, L and M shells, respectively, then the wavelength of $\text{K}_\alpha$ X-ray $(\lambda_1)$ is given by

$\lambda_1=\frac{\text{hc}}{\text{E}_\text{K}-\text{E}_\text{L}}$

Here,

h = Planck's constant

c = Speed of light

Wavelength of the $\text{K}_\beta$ x-ray $(\lambda_2)$ is given by

$\lambda_2=\frac{\text{hc}}{\text{E}_\text{K}-\text{E}_\text{M}}$

As the difference of energies (E_K - E_M) is more than (E_K - E_L), $\lambda_2$ is less than $\lambda_1$.

Hence, option (d) is not correct.

3. Will be halved.

Explanation:

Cut off wavelength is given by

$\lambda_{\text{min}}=\frac{\text{hc}}{\text{eV}'}$

where h = Planck's constan

c = speed of light

e = charge on an electron

V = potential difference applied to the tube

When potential difference (V) applied to the tube is doubled, cutoff wavelength

$\big({\lambda}'_\text{min}\big)$ is given by

${\lambda}'_\text{min}=\frac{\text{hc}}{\text{e}(2\text{V})}$

$\Rightarrow\lambda'_\text{min}=\frac{\lambda_\text{min}}{2}$

Cuttoff wavelength does not depend on the separation between the filament and the target.

Thus, cutoff wavelength will be halved if the the potential difference applied to the tube is doubled.

3. Will remain unchanged.

Explanation:

Cut off wavelength is given by

$\lambda_\text{min}=\frac{\text{hc}}{\text{eV}'}$

where h = Planck's constant

c = speed of light

e = charge on an electron

V = potential difference applied to the tube

Clearly, the cutoff wavelength does not depend on the current in the circuit and temperature of the filament.