Question

The threshold frequency of a metal is 10^15
Hz. When the metal is radiated with
radiations of frequencies 3 x 10^15 Hz and
5 x 10^15 Hz respectively, the ratio of
maximum kinetic energies of photo
electrons emitted is
(1) 3:5 (2) 5:3 (3) 1 : 2 (4) 2:1​

Answer 1

The correct option for the ratio of

The correct option for the ratio of maximum kinetic energies of photoelectrons emitted is (3) 1:2.

Given:

The threshold frequency of a metal is 10^15

Hz. When the metal is radiated with

radiations of frequencies 3 x 10^15 Hz and

5 x 10^15 Hz respectively.

To Find:

The ratio of maximum kinetic energies of photo

electrons emitted.

Solution:

To find the ratio of maximum kinetic energies of photoelectrons emitted we will follow the following steps:

As we know,

hv = hv0 + kinetic energy (KE)

Here, h is the constant and v0 is the threshold frequency.

v is the frequency of given radiation.

According to the question:

$vo = time es {10}^{15}$

$v1 \: = 3 \times {10}^{15}$

$v2 = 5 \times {10}^{15}$

So,

KE = h (v - v0)

Now,

Making two equations for the two conditions we get,

$KE(1) = h (3 \times {10}^{5} - 1 \times {10}^{5} ) \: eq - 1$

$KE(2) = h (5\times {10}^{5} - 1 \times {10}^{5} ) \: eq - 2$

Now,

Taking the ratio of two kinetic energies we get,

$\frac{KE(1)}{KE(2)} = \frac{h(3 \times {10}^{5} - 1 \times {10}^{5} )}{h(5 \times {10}^{5} - 1 \times {10}^{5} )} = \frac{2 \times {10}^{5} }{4 \times {10}^{} } = \frac{1}{2}$

Henceforth, the correct option for the ratio of maximum kinetic energies of photoelectrons maximum (3) is 1:2.

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