For photoelectric emission from certain metal the cutoff frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) :

Question

The variation of the stopping potential $ V_0 $ with the frequency $ \nu $ of the incident radiation for four metals A, B, C, and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for which metal?
variation of the stopping potential

Answer 1

To find the maximum possible velocity of the emitted electron given the photoelectric effect parameters, we need to understand and apply Einstein's photoelectric equation:

E = h \nu - \phi

where:

E is the kinetic energy of the emitted electron.
h is Planck's constant.
\nu is the frequency of the incident radiation.
\phi = h \nu_0 is the work function of the metal, with \nu_0 being the cutoff frequency.

According to the question, the cutoff frequency is \nu, and the incident radiation frequency is 2\nu. Therefore, the kinetic energy (K.E.) of the emitted electron can be calculated as:

K.E. = h(2\nu) - h\nu = h\nu

This kinetic energy can be expressed in terms of the electron's mass (m) and velocity (v) as follows:

K.E. = \frac{1}{2} m v^2

Setting the expressions for kinetic energy equal gives:

\frac{1}{2} m v^2 = h \nu

Solving for v, we get:

v^2 = \frac{2h\nu}{m}

v = \sqrt{\frac{2h\nu}{m}}

Thus, the maximum possible velocity of the emitted electron is given by:

\sqrt{\frac{2h\nu}{m}}

The correct answer is \sqrt{2h\nu/m}, which matches option $ \sqrt{2h v / m}$.

Answer 2