When a certain metal was irradiated with light of frequency $4.0 \times 10^{16} s^{- 1},$ the photo electrons emitted had three times the kinetic energy as the kinetic energy of photo electrons emitted when the metal was irradiated with light of frequency $2.0 \times 10^{16} s^{- 1} .$ Calculate the critical frequency $(V_{0})$ of the metal is $1 \times 10^{x} s^{- 1} .$ what is $' x' ?$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 16.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Using Einstein’s photo electric equation $\frac{K E_{2}}{K E_{1}} = \frac{h (υ_{2} - υ_{0})}{h (υ_{1} - υ_{0})} = \frac{υ_{2} - υ_{0}}{υ_{1} - υ_{0}}$

But given that $\frac{K E_{2}}{K E_{1}} = 3$

$\begin{array}{l} ∴ 3 = \frac{υ_{2} - υ_{0}}{υ_{1} - υ_{0}} \Rightarrow 3 (υ_{1} - υ_{0}) = υ_{2} - υ_{0} \\ \Rightarrow 3 υ_{1} - υ_{2} = 3 υ_{0} - υ_{0} = 2 υ_{0} \\ \Rightarrow υ_{0} = \frac{2 \times 10^{16}}{2} = 1 \times 10^{16} s^{- 1} \end{array}$