Question

What is the threshold frequency of a metal whose work
function is 13.26 x 10-25 J?
(h=6.63 * 10-34 Js)
• 9*10-8 Hz
1x1010 Hz
2 x 100 Hz
5* 100 Hz​

Answer 1

Answer:-

$\red{\bigstar}$ Threshold frequency

$\large\leadsto\boxed{\sf \green{2 \times 10^{-9}Hz}}$

• Given:-

Work function = $\bf{13.26 \times 10^{-25}J}$

• To Find:-

Threshold frequency = ?

• Solution:-

$\pink{\bigstar}$$\boxed{\sf{Work \: function [W_{0}] = Planck's \: Constant [h] \times Threshold \: frequency [v_{0}]}}$

here,

h [Planck's constant] = 6.63 × 10 -³⁴ Js

Hence,

→ $\sf{13.26 \times 10^{-25} J = 6.63 \times 10^{-34} Js \times v_{0}}$

→ $\sf{v_{0} = \dfrac{13.26 \times 10^{-25}J}{6.63 \times 10^{-34}Js}}$

→ $\sf \green{v_{0} = 2 \times 10^{-9}sec^{-1}}$

Therefore, the threshold frequency of the metal is $\bf{2 \times 10^{-9}Hz}$