Question

For photoelectric emission from certain metal the cut-off 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)

• A. $\sqrt{\frac{h}{2m}}$
• B. $\sqrt{\frac{hv}{m}}$
• C. $\sqrt{\frac{2hv}{m}}$
• D. $2\sqrt{\frac{hv}{m}}$

Answer 1

The Correct Option is C

The question involves understanding the concept of photoelectric emission, where electrons are emitted from a metal surface when it is exposed to electromagnetic radiation above a certain frequency known as the cut-off frequency.

The formula describing this effect, based on Einstein’s photoelectric equation, is:

K.E. = h(f - f_0)

where:

• K.E. is the kinetic energy of the emitted electron.
• h is Planck's constant.
• f is the frequency of the incident radiation.
• f_0 is the cut-off frequency or threshold frequency.

Given in the problem:

• Cut-off frequency, f_0 = v.
• Incident frequency, f = 2v.

Substituting into Einstein's equation, the kinetic energy becomes:

K.E. = h(2v - v) = hv

Since kinetic energy is also given by \frac{1}{2}mv^2 (where v here is the velocity of the electron), we can equate and solve for the velocity:

\frac{1}{2}mv_{\text{electron}}^2 = hv

Rearranging, we find:

v_{\text{electron}}^2 = \frac{2hv}{m}

Taking the square root of both sides gives:

v_{\text{electron}} = \sqrt{\frac{2hv}{m}}

Therefore, the maximum possible velocity of the emitted electron is given by the expression \sqrt{\frac{2hv}{m}}. Thus, the correct answer is \sqrt{\frac{2hv}{m}}.

Let's confirm why the other options are incorrect:

• \sqrt{\frac{h}{2m}}: This doesn't consider the frequency hv factor, which is critical for the energy imparted to the electron.
• \sqrt{\frac{hv}{m}}: This accounts for the frequency but misses the factor of 2, a necessary factor given the derived equation.
• 2\sqrt{\frac{hv}{m}}: This exaggerates the factor within the radical, leading to an incorrect estimation velocity.