Question

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other, the maximum velocities of the photoelectrons emitted in the two cases are \( V_1 \) and \( V_2 \) respectively. The ratio \( V_1 : V_2 \) is

• A. \( \sqrt{3} : \sqrt{2} \)
• B. \( \sqrt{2} : 1 \)
• C. \( \sqrt{3} : 1 \)
• D. \( 1 : \sqrt{2} \)

Hint:

In the photoelectric effect, the kinetic energy of the ejected electrons depends on the frequency of the incident photons. If the frequency increases, the kinetic energy and velocity of the electrons increase accordingly.

Answer 1

The Correct Option is A

Step 1: Understanding the photoelectric effect.
The photoelectric effect is the phenomenon in which electrons are ejected from the surface of a material when it is exposed to light (photons). The energy of the photons must be greater than the work function \( W \) of the material for electrons to be ejected. The kinetic energy of the ejected electrons is given by the photoelectric equation:
\[ K.E. = h\nu - W, \]
where:
- \( h \) is Planck’s constant,
- \( \nu \) is the frequency of the incident photons,
- \( W \) is the work function of the metal.
The maximum velocity \( v \) of the photoelectrons is related to the kinetic energy by:
\[ K.E. = \frac{1}{2} m v^2, \] where \( m \) is the mass of the ejected electron.

Step 2: Relating the velocity to the frequency of the incident photon.
From the photoelectric equation, we have:
\[ h\nu - W = \frac{1}{2} m v^2. \]
Now, let’s denote the frequency of the photons as \( \nu_1 \) for the case when the photon energy is twice the work function, and \( \nu_2 \) for the case when the photon energy is three times the work function. Thus:
\[ K.E. = \frac{1}{2} m v^2 = h\nu - W. \]
For the first case (\( \nu_1 = 2W \)):
\[ K.E._1 = h \times 2W - W = W, \]
so the velocity is: \[ v_1 = \sqrt{\frac{2W}{m}}. \]
For the second case (\( \nu_2 = 3W \)):
\[ K.E._2 = h \times 3W - W = 2W, \]
so the velocity is: \[ v_2 = \sqrt{\frac{4W}{m}}. \]

Step 3: Finding the ratio of the velocities.
The ratio of the velocities \( v_1 \) and \( v_2 \) is:
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{2W}{m}}}{\sqrt{\frac{4W}{m}}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}. \]
Thus, the ratio \( V_1 : V_2 \) is \( \sqrt{3} : \sqrt{2} \).
Final Answer:
The ratio \( V_1 : V_2 \) is:
\[ \boxed{\sqrt{3} : \sqrt{2}}. \]