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. $6\times 10^{15}$

Answer 1

Answer: (B)

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

$h\left(1.6\times 10^{16}-v_0\right)=2K.E⟶$ Eq .(i)

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

$h\left(1.0\times 10^{16}-v_0\right)=K.E⟶$ (Eq .(ii))

From Eqs. (i) and (ii)

$\frac{h\left(1.6\times 10^{16}-v_0\right)}{h\left(1.0\times 10^{16}-v_0\right)}=\frac{2K\cdot E}{K.E}$

$1.6\times 10^{16}-v_0=2\times 1.0\times 10^{16}-v_0$

$1.6\times 10^{16}-2\left(1.0\times 10^{16}\right)=4\times 10^{15}Hz$

$v_0=4\times 10^{15}Hz$