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Calculate the maximum velocity of photoelectrons if $\upsilon = 2{\upsilon _0}$ and for a certain metal $v$ is five times the ${\upsilon _0}$ when the maximum velocity of coming photo – electrons is $8 \times {10^6}$.
A) $4 \times {10^6}m/s$
B) $6 \times {10^6}m/s$
C) $8 \times {10^6}m/s$
D) $1 \times {10^6}m/s$

Last updated date: 26th Feb 2024
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Hint: Use the formula of maximum kinetic energy. Then, put the respective values in that expression, once using the initial frequency and then using the final frequency. Next, divide both these equations and find the value of velocity.

Complete step by step answer:
Let the frequency of excitation be ${\upsilon _0}$.
Then, according to the question, it is given that –
Initial frequency, $\upsilon = 5{\upsilon _0}$
Let the initial maximum velocity of photo – electrons be ${v_1}$
$\therefore {v_1} = 8 \times {10^6}m/s$
Now, using the maximum kinetic energy formula –
$\dfrac{1}{2}m{v^2} = h\left( {\upsilon - {\upsilon _0}} \right)$
Therefore, putting the values from question in the above formula –
 \implies \dfrac{1}{2}mv_1^2 = h\left( {5{\upsilon _0} - {\upsilon _0}} \right) \\
 \implies \dfrac{1}{2}m \times {\left( {8 \times {{10}^6}} \right)^2} = 4{\upsilon _0}h \cdots \left( 1 \right) \\
Now, for the second equation –
According to the question it is given that –
Final frequency, $\upsilon = 2{\upsilon _0}$
Let the final maximum velocity of the photo – electrons be ${v_2}$.
Again, using the maximum kinetic energy formula –
$\dfrac{1}{2}mv_2^2 = h\left( {\upsilon - {\upsilon _0}} \right)$
Putting the values from the question in the above formula –
 \implies \dfrac{1}{2}mv_2^2 = h\left( {2{\upsilon _0} - {\upsilon _0}} \right) \\
 \implies \dfrac{1}{2}mv_2^2 = h{\upsilon _0} \cdots \left( 2 \right) \\
Now, dividing the equation $\left( 2 \right)$ by equation $\left( 1 \right)$ -
After cancelling $m$ and $\dfrac{1}{2}$ on both the equations, we get –
$\implies \dfrac{{v_2^2}}{{{{\left( {8 \times {{10}^6}} \right)}^2}}} = \dfrac{{h{\upsilon _0}}}{{4{\upsilon _0}h}}$
Cancelling $h$ and ${\upsilon _0}$ on both numerator and denominator, we get –
$\implies \dfrac{{v_2^2}}{{{{\left( {8 \times {{10}^6}} \right)}^2}}} = \dfrac{1}{4}$
By cross – multiplication, we get –
$v_2^2 = \dfrac{{{{\left( {8 \times {{10}^6}} \right)}^2}}}{4}$
Finding square root on both sides, we get –
  \implies {v_2} = \dfrac{{8 \times {{10}^6}}}{2} \\
  \implies {v_2} = 4 \times {10^6}m/\sec \\
Hence, the maximum velocity of the photo – electrons is $4 \times {10^6}m/\sec $.

Therefore, the correct option is (A).

Note: Kinetic energy of the photo – electron increases linearly with $\upsilon $ as long as the photon energy is greater than the work function. The kinetic energy of photoelectrons increases with the increase in frequency of light and remains constant when the amplitude of light increases.
The required energy by a particular metal to free the electron from it is called the work function of that metal.