Question 260: Dual Nature of Matter and Radiation

Question

Energy of the incident photons on the metal surface is initially 4 W and then 6 W where $W$ is the work function of that metal. The ratio of velocities of emitted photoelectrons is

Accepted Answer

Step 1: Recall photoelectric equation
$ K_{	ext{max}} = E_{	ext{photon}} - W $
and
$ K_{	ext{max}} = 	frac{1}{2} m v^2 $
So,
$ v \propto \sqrt{E_{	ext{photon}} - W} $
Step 2: Photon energies
Case 1: $E_1 = 4W$ Ã¢ÂÂ
$K_1 = 4W - W = 3W$
Case 2: $E_2 = 6W$ Ã¢ÂÂ
$K_2 = 6W - W = 5W$
Step 3: Velocities
$ v_1 \propto \sqrt{3W},  \quad v_2 \propto \sqrt{5W} $
So the ratio is
$ v_1 : v_2 = \sqrt{3W} : \sqrt{5W} = \sqrt{3} : \sqrt{5}. $
Ã¢ÂÂ Final Answer:
Option A: $\sqrt{3} : \sqrt{5}$

Answer

$\sqrt{3}: \sqrt{5}$

Answer

$1: 2$

Answer

$2: 3$

Answer

$\sqrt{2}: \sqrt{3}$

Dual Nature of Matter and Radiation

Energy of the incident photons on the metal surface is initially 4 W and then 6 W where W is the work function of that metal. The ratio of velocities of emitted photoelectrons is

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