Question

threshold frequency of a surface is v. It is illuminated by 3 v, frequency, then maximum speed of photo
electrons is V m/sec. What will be maximum speed if incident frequency is 9 Vo?
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Answer 1

It Will be $\sqrt 3$ times its Original Value

Explanation:

• Let $\nu_1$ be the incident frequency, then we write the photoelectric equation as $h\nu_1 = \phi + (\frac {1}{2})m\nu^2$ ....................( 1 )

 

• where $\nu$ is the velocity of electron that corresponds to maximum kinetic energy, $\phi$ is work function and h is Planck's constant.

 

• if the incident frequency becomes $3 \nu_1$, photon energy becomes 3 times of original energy and (1) can be written as

 

$h(3 \nu_1) = \phi + \Delta E +(\frac {1}{2})m(\sqrt 3 \nu)^2$ ................(2)

 

In the above equation, the term $\Delta E$ should be absorbed in the term representing maximum kinetic energy of electron.

 

Hence  $h(3 \nu_1) = \phi + (\frac {1}{2})m (\sqrt 3\nu + \Delta \nu)^2$ ................(3)

 

Hence from (3), it can be concluded that when frequency is changed to thrice of its original value, maximum velocity of photo-electron will be grater than $\sqrt3$ times of its original value.

 

(I guess $\sqrt$ is not typed while typing the various options for answer of the question, because we need exact values of $\phi\ and\ \nu_1$ to compare the maximum velocity with 3 times of original maximum velocity V)