Question

A certain metal when irradiated with light $(\nu =3.2\times {10}^{16}Hz)$ emits photoelectrons with twice kinetic energy of photoelectrons that are emitted when the same metal is irradiated by light $(\nu =2.0\times {10}^{16}Hz)$. Calculate ${\nu }_o$ of electron.

• A. $1.2\times {10}^{14}Hz$
• B. $8\times {10}^{15}Hz$
• C. $1.2\times {10}^{16}Hz$
• D. $4\times {10}^{12}Hz$

Answer 1

The correct option is B $8\times {10}^{15}Hz$

Applying Einstein's Photoelectric Equation,
$K.E=E-W_o$
$K.E=h\nu -h{\nu }_o$
In first case,$K.E_1=h(3.2\times {10}^{16}-{\nu }_o)$
In second case,$K.E_2=h(2.0\times {10}^{16}-{\nu }_o)$
It is given in that $(K.E_1)=2(K.E_2)$
Hence,$\left[h\right(3.2\times {10}^{16}-{\nu }_o\left)\right]=2\left[h\right(2.0\times {10}^{16}-{\nu }_o\left)\right]$
On solving, ${\nu }_o=8\times {10}^{15}Hz$