Question

When a certain metal surface is illuminated with light of frequency $ \nu $, the stopping potential for photoelectric current is $ V_0 $. When the same surface is illuminated by light of frequency $ \frac{\nu}{2} $, the stopping potential is $ \frac{V_0}{4} $. The threshold frequency for photoelectric emission is

• A. \( \frac{\nu}{2} \)
• B. \( \frac{\nu}{3} \)
• C. \( \frac{3\nu}{4} \)
• D. \( \nu \)

Hint:

The threshold frequency for photoelectric emission is the frequency at which the stopping potential becomes zero. Use the photoelectric equation to solve for the threshold frequency in terms of the given frequencies.

Answer 1

The Correct Option is B

The photoelectric effect is governed by Einstein’s photoelectric equation:

\[ eV_0 = h\nu - h\nu_0 \]

where \( e \) is the electron charge, \( V_0 \) is the stopping potential, \( h \) is Planck’s constant, \( \nu \) is the light frequency, and \( \nu_0 \) is the threshold frequency.

Given:

• For frequency \( \nu \), stopping potential \( = V_0 \)
• For frequency \( \frac{\nu}{2} \), stopping potential \( = \frac{V_0}{4} \)

From \( eV_0 = h\nu - h\nu_0 \), for frequency \( \nu \):

\[ eV_0 = h\nu - h\nu_0 \quad ...(1) \]

For frequency \( \frac{\nu}{2} \), we have:

\[ e\frac{V_0}{4} = h\left(\frac{\nu}{2}\right) - h\nu_0 \quad ...(2) \]

Let's solve equations (1) and (2). Multiply equation (2) by 4:

\[ eV_0 = 2h\left(\frac{\nu}{2}\right) - 4h\nu_0 \]

Simplifying gives:

\[ eV_0 = h\nu - 4h\nu_0 \quad ...(3) \]

We equate equations (1) and (3):

\[ h\nu - h\nu_0 = h\nu - 4h\nu_0 \]

Cancel \( h\nu \) and divide by \( h \):

\[ -\nu_0 = -4\nu_0 \]

Thus, solving for \( \nu_0 \) gives:

\[ 3\nu_0 = \nu \]

Therefore,

\[ \nu_0 = \frac{\nu}{3} \]

The threshold frequency \( \nu_0 \) for photoelectric emission is \( \frac{\nu}{3} \).

Answer 2

To find the threshold frequency \( \nu_0 \), we use the photoelectric equation: \( eV_0 = h(\nu - \nu_0) \).

1. For frequency \( \nu \), the stopping potential is \( V_0 \):
\( eV_0 = h(\nu - \nu_0) \).

2. For frequency \( \frac{\nu}{2} \), the stopping potential is \( \frac{V_0}{4} \):
\( e\left(\frac{V_0}{4}\right) = h\left(\frac{\nu}{2} - \nu_0\right) \).

Now, solve these equations simultaneously:

From equation 1:
\( eV_0 = h(\nu - \nu_0) \) ⇒ \( \nu_0 = \nu - \frac{eV_0}{h} \).

From equation 2:
\( e\left(\frac{V_0}{4}\right) = h\left(\frac{\nu}{2} - \nu_0\right) \)
⇒ \( \frac{eV_0}{4h} = \frac{\nu}{2} - \nu_0 \).

Substitute \( \nu_0 = \nu - \frac{eV_0}{h} \) into the second equation:

\( \frac{eV_0}{4h} = \frac{\nu}{2} - \left(\nu - \frac{eV_0}{h}\right) \).

Simplify:
\( \frac{eV_0}{4h} = \frac{\nu}{2} - \nu + \frac{eV_0}{h} \)
⇒ \( \frac{eV_0}{4h} = -\frac{\nu}{2} + \frac{eV_0}{h} \).

Rearrange:
\( \frac{eV_0}{4h} + \frac{\nu}{2} = \frac{eV_0}{h} \).

Calculate:
\( \frac{\nu}{2} = \frac{3eV_0}{4h} \)

Plug this back into \( \nu_0 = \nu - \frac{eV_0}{h} \):
Using \( \frac{\nu}{2} = \frac{3eV_0}{4h} \), so, \( \nu = \frac{3eV_0}{2h} \)
⇒ \( \nu_0 = \frac{3eV_0}{2h} - \frac{eV_0}{h} = \frac{3eV_0}{2h} - \frac{2eV_0}{2h} \)
⇒ \( \nu_0 = \frac{eV_0}{2h} \).

Thus, comparing \( \nu_0 \) in terms of \( \nu = \frac{3eV_0}{2h} \):
⇒ \( \nu_0 = \frac{\nu}{3} \).

The threshold frequency for the photoelectric emission is \( \frac{\nu}{3} \).