Question

A light of frequency $1.6 \times 10^{16} \, Hz$ when falls on a metal plate emits electrons that have double the kinetic energy compared to the kinetic energy of emitted electrons when frequency of $1.0 \times 10^{16} \, Hz$ falls on the same plate. The threshold frequency $(v_0)$ of the metal in $Hz$ is

• A. $1 \times 10^{15}$
• B. $4 \times 10^{15}$
• C. $3 \times 10^{15}$
• D. $4 \times 10^{15}$

Answer 1

The Correct Option is B

When $1.6 \times 10^{16}$ Hz frequency falls on metal plate than double kinetic energy obtained.

$h\left(1.6 \times 10^{16}-v_{0}\right)=2 K.E \longrightarrow$ Eq .(i)

When $1.0 \times 10^{16}\, Hz$ frequency falls on same plate than K.E.

$h\left(1.0 \times 10^{16}-v_{0}\right)= K.E \longrightarrow $ (Eq .(ii))

From Eqs. (i) and (ii)

$\frac{h\left(1.6 \times 10^{16}-v_{0}\right)}{h\left(1.0 \times 10^{16}-v_{0}\right)}=\frac{2 K \cdot E }{ K . E }$
$1.6 \times 10^{16}-v_{0}=2 \times 1.0 \times 10^{16}-v_{0}$
$1.6 \times 10^{16}-2\left(1.0 \times 10^{16}\right)=4 \times 10^{15}\,Hz$
$v_{0}=4 \times 10^{15}\, Hz$