Question

Figure shows the variation of frequency of a characteristic X-ray and atomic number.

• A. The characteristic X-ray is $K_{\alpha}$
• B. The characteristic X-ray is $K_{\beta}$
• C. The energy of photon emitted when this X-ray is emitted by a metal having z = 101 is 204 keV
• D. The energy of photon emitted when this X-ray is emitted by a metal having z = 101 is 102 keV

Answer 1

Answer: (A), (D)

The characteristic X-ray is $K_{\alpha}$, The energy of photon emitted when this X-ray is emitted by a metal having z = 101 is 102 keV

Answer 2

The graph shows the variation of frequency $\nu$ of a characteristic X-ray with atomic number $Z$. According to Moseley's law, for a characteristic X-ray of a given series, the frequency is related to the atomic number by $\sqrt{\nu} = a(Z-b)$, where $a$ and $b$ are constants. Thus, $\nu = a^2 (Z-b)^2$. For K-series X-rays, the screening constant $b$ is approximately 1. The graph shows that $\nu \approx 0$ when $Z=1$, which is consistent with $b=1$. So we can write $\nu = A(Z-1)^2$ for some constant $A$.

The energy of the emitted photon is $E = h\nu = hA(Z-1)^2$. Let $E_0 = hA$. Then $E = E_0 (Z-1)^2$. The constant $E_0$ depends on the specific transition. For the K-series, the transitions are to the K shell ($n_1=1$).

For the $K_{\alpha}$ line, the transition is from the L shell ($n_2=2$) to the K shell ($n_1=1$). The energy is given by $E_{K_{\alpha}} = R_y (Z-1)^2 (\frac{1}{1^2} - \frac{1}{2^2}) = R_y (Z-1)^2 \frac{3}{4}$. So, $E_0 = R_y \frac{3}{4}$, where $R_y$ is the Rydberg constant for energy, approximately 13.6 eV. $E_0 \approx 13.6 \times \frac{3}{4} = 10.2$ eV. Thus, $E_{K_{\alpha}} \approx 10.2 (Z-1)^2$ eV.

For the $K_{\beta}$ line, the transition is from the M shell ($n_2=3$) to the K shell ($n_1=1$). The energy is given by $E_{K_{\beta}} = R_y (Z-1)^2 (\frac{1}{1^2} - \frac{1}{3^2}) = R_y (Z-1)^2 \frac{8}{9}$. So, $E_0 = R_y \frac{8}{9}$. $E_0 \approx 13.6 \times \frac{8}{9} \approx 12.09$ eV. Thus, $E_{K_{\beta}} \approx 12.09 (Z-1)^2$ eV.

Now let's evaluate the energy for $Z=101$. If the characteristic X-ray is $K_{\alpha}$, $E_{K_{\alpha}}(Z=101) \approx 10.2 (101-1)^2 = 10.2 \times 100^2 = 10.2 \times 10000 = 102000$ eV = 102 keV. If the characteristic X-ray is $K_{\beta}$, $E_{K_{\beta}}(Z=101) \approx 12.09 (101-1)^2 = 12.09 \times 10000 = 120900$ eV = 120.9 keV.

The calculated energy for $K_{\alpha}$ at Z=101 is 102 keV, which matches the fourth option. This suggests that the characteristic X-ray is $K_{\alpha}$.