Two identical photocathodes receive light of frequencies ' n 1 ' and ' n 2 '. If the velocities of the emitted photoelectrons of mass ' m ' are ' V 1 ' an

Question

Two identical photocathodes receive light of frequencies ' n 1 ​ ' and ' n 2 ​ '. If the velocities of the emitted photoelectrons of mass ' m ' are ' V 1 ​ ' and ' V 2 ​ ' respectively, then ( h = Planck's constant )

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by:

$\frac{1}{2} m V^{2} = hn - Φ$

where $n$ is the frequency of incident light and $Φ$ is the work function of the material.

For the first photocathode:

$\frac{1}{2} m V_{1}^{2} = h n_{1} - Φ$

For the second identical photocathode (same work function $Φ$ ):

$\frac{1}{2} m V_{2}^{2} = h n_{2} - Φ$

Subtracting the second equation from the first, we get:

$\frac{1}{2} m V_{1}^{2} - \frac{1}{2} m V_{2}^{2} = h n_{1} - Φ - (h n_{2} - Φ)$

$\frac{1}{2} m (V_{1}^{2} - V_{2}^{2}) = h (n_{1} - n_{2})$

$V_{1}^{2} - V_{2}^{2} = \frac{2 h}{m} (n_{1} - n_{2})$

Answer: $V_{1}^{2} - V_{2}^{2} = \frac{2 h}{m} (n_{1} - n_{2})$

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