Question

Light of wavelength $4000 A^0$ is incident on a metal surface of work function $2.5\, e V$. Given $h=6.62 \times 10^{-34} \,Js,\, c = 3 \times 10^8 m/s,$ the maximum KE of photoelectrons emitted and the corresponding stopping potential are respectively

• A. 0. 6 eV, 0.6 V
• B. 2.5 eV, 2.5 V
• C. 3.1 eV,3.1 V
• D. 0.6 eV, 0.3 V

Answer 1

Answer: (A)

According to given question, Maximum kinetic energy is $K$. $E$.,
$
\begin{array}{l}
\frac{ hC }{\lambda}- W = K . E . \\
\Rightarrow K . E .=4.14 \times 10^{-15} \times 3 \times 10^{8}-2.5 \\
\Rightarrow K . E .=0.605 eV \\
\Rightarrow K . E . \approx 0.6 eV
\end{array}
$
So,
From definition of stopping potential,
K.E. $= e \times V _{0}$ [Where $V _{0}$ is stopping potential]
$0.6 eV = eV _{0}$
$\Rightarrow V _{0}=0.6 volts$