Q156 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Physics
A monochromatic light of wavelength $\lambda$ ejects photoelectrons from a metal surface with work function $(\phi) 2.4 \mathrm{eV}$. These photoelectrons are made to collide with hydrogen atoms in ground state. The maximum value of $\lambda$ for which hydrogen atom may be ionised is [take, $h c=1240 \mathrm{eV}-\mathrm{nm}$ ]
- A$80 \mathrm{~nm}$
- Correct answerB$77.5 \mathrm{~nm}$
- C$75.5 \mathrm{~nm}$
- D$85 \mathrm{~nm}$
Solution
Given, work-function, $\phi=2.4 \mathrm{eV}$
Energy required to ionise $\mathrm{H}$-atom in ground state
$=13.6 \mathrm{eV}$
According to photoelectric equation,
$E=\mathrm{KE}+\phi \Rightarrow \frac{h c}{\lambda}=\mathrm{KE}+\phi$
$\Rightarrow \quad \frac{1240}{\lambda}=13.6+2.4 \quad(\because h c=1240 \mathrm{eV}-\mathrm{nm})$
$\Rightarrow \quad \frac{1240}{\lambda}=16 \quad \therefore \quad \lambda=\frac{1240}{16}=77.5 \mathrm{~nm}$
Hence, the correct answer is option: 'B'.