Question

A monochromatic light of wavelength $\lambda$ ejects photoelectrons from a metal surface with work function (\varphi) $2.4\, eV$. These photoelectrons are made to collide with hydrogen atoms in ground state. The maximum value of $\lambda$ for which hydrogen atom may be ionized is [Take hc=1240\, eV-nm]

• A. 80 nm
• B. 77.5 nm
• C. 75.5 nm
• D. 85 nm

Answer 1

Answer: (B)

Given, work-function, $\phi=2.4\, eV$
Energy required to ionise $H$-atom in ground state
$=13.6\, eV$
According to photoelectric equation,
$E=K E+\varphi \Rightarrow \frac{h c}{\lambda}=K E+\varphi$
$\Rightarrow \frac{1240}{\lambda}=13.6+2.4$
$(\because h c=1240\, eV - nm )$
$\Rightarrow \frac{1240}{\lambda}=16$
$\therefore \lambda=\frac{1240}{16}$
$=77.5\, nm$