A beam of monochromatic light of wavelength ejects photoelectrons… - Vidyadip

Question

A beam of monochromatic light of wavelength $\lambda$ ejects photoelectrons from a cesium surface $\big(\Phi=1.9\text{eV}\big).$ These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of $\lambda$ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.

✓

Answer

$\phi=1.9\text{eV}$

The hydrogen is ionized,

$\text{n}_1=1,\text{n}_2=\infty$

Energy required for ionization $=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)=13.6$

$\frac{\text{hc}}{\lambda}-1.9=13.6$

$\lambda=80.1\text{nm}=80\text{nm}$

For the electron to be excited from,

$\text{n}_1\text{ to n}_2=2$

$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)=13.6\Big(1-\frac{1}{4}\Big)=\frac{13.6\times3}{4}$

$\frac{\text{hc}}{\lambda}-1.9=\frac{13.6\times3}{4}$

$\lambda=\frac{1242}{12.1}=102.64=102\text{nm}$

$\text{E}_2=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)$

$\text{E}_2=13.6\Big(\frac{1}{4}-\frac{1}{9}\Big)$

$\text{E}_2=\frac{13.66\times5}{36}$

For Einstein's photoelecrtic equation,

$\frac{\text{hc}}{\lambda}-1.9=\frac{13.6\times5}{36}$

$\frac{\text{hc}}{\lambda}=\frac{13.6\times5}{36}+1.9$

$\frac{1240}{\lambda1}=1.88+1.9=3.78$

$\lambda=\frac{1240}{3.78}$

$\lambda=328.04\text{nm}$

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