Question

A certain metal when irradiated with light 3.2×10^16 Hz emits photo electrons twice KE

Answer 1

Explanation:

E(Total Energy) = $E_0$(Minimum Energy required for the electron) + K.E(kinetic Energy of Photo electron)

$h v=h v_{0}+\frac{1}{2} m v^{2}$

$6.6 \times 10^{-34} \times\left(3.2 \times 10^{16}\right)=E_{0}+K . E$

$2.112 \times 10^{-17}=E_{0}+K . E \rightarrow(1)$

$6.6 \times 10^{-34} \times\left(2 \times 10^{16}\right)=E_{0}+2 K . E$

$1.32 \times 10^{-17}=E_{0}+2 K . E \rightarrow(2)$

On solving, equation (1) and (2), we get,

$K . E=-7.92 \times 10^{-18} \ \mathrm{J}$

On substituting, kinetic energy in equation (1), we get,

$2.112 \times 10^{-17}=E_{0}-7.92 \times 10^{-18}$

Thus, the minimum energy is,

$E_{0}=2.904 \times 10^{-17} \ \mathrm{J}$

Now, the total energy is,

$E=\left(2.904 \times 10^{-17}\right)+\left(-7.92 \times 10^{-18}\right)$

$\Rightarrow E=2.112 \times 10^{-17} \ \mathrm{J}$

Thus, now the threshold frequency is,

$E_{0}=h v_{0}$

On substituting the found values, we get,

$\Rightarrow 2.904 \times 10^{-17}=\left(6.6 \times 10^{-34}\right) v_{0}$

$v_{0}=\frac{2.904 \times 10^{-17}}{6.6 \times 10^{-34}}$

$\therefore v_{0}=4.4 \times 10^{16} \ H z$

Answer 2

Explanation:

hii. .I think the answer will help you...