Question

if a light with frequency 4×10^16 Hz emitted photoelectron with double the maximum kinetic energy is emitted by the light of frequency 2.5×10^16 from the same metal surface then what is threshold frequency of metal ???

Answer 1

your answer mate l..........

Answer 2

Threshold frequency of metal is $8 \times 10^{15} \mathrm{Hz}$.

Given:  

Initial frequency$\mathrm{v}=4 \times 10^{16} \mathrm{HZ}$

Final frequency $V_{0}=2.5 \times 10^{16} H Z$

To find:

The threshold frequency for the given metal.

Solution:

As per the given question, a frequency of light emitted photo electron with the “maximum kinetic energy” doubled by the light on the ‘metal surface’.

Formula for initial Kinetic energy,

$K E_{1}=h\left(V_{1}-V_{0}\right) \ldots \ldots \ldots \ldots(i)$

Formula for Final kinetic energy,  

$K E_{2}=h\left(V_{2}-V_{0}\right) \ldots \ldots \ldots \ldots .(i i)$

Dividing the equation (i) and (ii)

After dividing we get,  

$\frac{V_{2}-V_{0}}{V_{1}-V_{0}}=\frac{1}{2}$

$\frac{2.5 \times 10^{16}-V_{0}}{4 \times 10^{16}-V_{0}}=\frac{1}{2}$

$5 \times 10^{16}-2 V_{0}=4 \times 10^{16}-V$

$V_{o}=1 \times 10^{16} H z$

$V_{o}=1 \times 10^{16} H z$

Therefore, Threshold frequency of metal is $1 \times 10^{16} \mathrm{Hz}$.