Question

A certain metal when irradiated with light $(v=3.2\times 10^{16}Hz$ ) emits photo electrons with twice kinetic energy as did photo electrons when the same metal is irradiated by light $\left(v=2.0\times 10^{16}Hz\right)$. Calculate $v_0$ 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

Answer: (B)

K. E. $=h\left(v-v_0\right)$
K.E. of photoelectrons when
$v=3.2\times 10^{16}Hz$
$K.E_1=h\left(3.2\times 10^{16}-v_0\right)$
$K$. E. of photoelectron when
$v=2.0\times 10^{16}Hz$
K. $E_2=h\left(2.0\times 10^{16}-v_0\right)$
According to question $K.E_1=2K$. $E_2$
$\therefore h\left(3.2\times 10^{16}-v_0\right)=2h\left(2.0\times 10^{16^2}-v_0\right)$
$3.2\times 10^{16}-v_0=4.0\times 10^{16}-2v_0$
$v_0=4.0\times 10^{16}-3.2\times 10^{16}$
$=0.8\times 10^{16}Hz=8\times 10^{15}Hz$
$=8\times 10^{15}Hz$