The maximum velocity of electrons emitted fro | vedclass

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(B) According to Einstein's photoelectric equation: $hf = W_0 + \frac{1}{2}mv^2$. Since $v = V$ for frequency $f$,we have $V = \sqrt{\frac{2(hf - W_0)

Vedclass · Accepted Answer

$> 2V$

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(B) According to Einstein's photoelectric equation: $hf = W_0 + \frac{1}{2}mv^2$. Since $v = V$ for frequency $f$,we have $V = \sqrt{\frac{2(hf - W_0)}{m}}$. When the frequency becomes $4f$,the new maximum velocity $V'$ is given by: $V' = \sqrt{\frac{2(h(4f) - W_0)}{m}}$. We can rewrite this as: $V' = \sqrt{\frac{2(4hf - 4W_0 + 3W_0)}{m}} = \sqrt{4 \left( \frac{2(hf - W_0)}{m} \right) + \frac{6W_0}{m}}$. Since $V = \sqrt{\frac{2(hf - W_0)}{m}}$,we get $V' = \sqrt{4V^2 + \frac{6W_0}{m}}$. Since $\frac{6W_0}{m} > 0$,it follows that $V' > \sqrt{4V^2}$,which means $V' > 2V$.

The maximum velocity of electrons emitted from a metal surface is $V$ when the frequency of light falling on it is $f$. What is the maximum velocity when the frequency becomes $4f$?

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