Question

Light of frequency 7.21 X 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 X 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Answer 1

Answer:

The threshold frequency for photo emission of electrons$v' = 4.74 \times 10^{14} Hz$.

Explanation:

As per the question,

Given data:

$Light \ of \ frequency = 7.21 \times 10^{14}Hz$

$Maximum \ speed = 6.0\times 10^{5} \frac{m}{s}$

Formula used:

According to Photoelectric equation:

Maximum K. E = hv – hv'

Where,

h = plank’s constant

$h=6.634 \times 10^{-34} \frac{m^{2}kg}{s}$

v = frequency of photon

v' = threshold frequency

Also,

We know that maximum kinetic energy of photo-electrons is also given by:

$K. E = \frac{1}{2}mv_{0}^{2}$

where,

$v_{0} = 6 \times 10^{5} \frac{m}{s^{2}}$

Therefore we can say that,

$hv - hv' = \frac{1}{2} mv_{0}^{2}$

$v' = v + \frac{1}{h} \times \frac{1}{2}mv_{0}^{2}$

By putting all the values and further solving, we get,

$v' = 4.74 \times 10^{14} Hz$

Hence, the threshold frequency for photo emission of electrons$v' = 4.74 \times 10^{14} Hz$.