Question 373: Dual Nature of Matter and Radiation

Question

Energy of the incident photon on the metal surface is $$3 W$$ and then $$5 W$$, where $$W$$ is the work function for that metal. The ratio of velocities of emitted photoelectrons is

Accepted Answer

To find the ratio of velocities of the emitted photoelectrons when the energy of the incident photon is $$3 W$$ and then $$5 W$$, where $$W$$ is the work function of the metal, we can use the photoelectric equation. The kinetic energy (KE) of the emitted photoelectron can be described by the equation:
$KE = h
u - W$
where $h
u$ is the energy of the incident photon and $W$ is the work function of the metal.
For two different incident photon energies $3W$ and $5W$, let's denote the kinetic energies of the emitted photoelectrons as $KE_1$ and $KE_2$, respectively:
$KE_1 = 3W - W = 2W$
$KE_2 = 5W - W = 4W$
The kinetic energy is also given by the equation $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the photoelectron and $v$ is its velocity. Thus, for the two cases above, we have:
$\frac{1}{2}mv_1^2 = 2W$
$\frac{1}{2}mv_2^2 = 4W$
Dividing the second equation by the first gives:
$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4W}{2W}$
Simplifying this, we get:
$\frac{v_2^2}{v_1^2} = 2$
Taking the square root of both sides to find the ratio of the velocities $v_2:v_1$, we find:
$\frac{v_2}{v_1} = \sqrt{2}$
Therefore, the correct answer is $1: \sqrt{2}$, which corresponds to Option C.

Answer

$$1: 4$$

Answer

$$1: 2$$

Answer

$$1: \sqrt{2}$$

Answer

$$1: 1$$

Dual Nature of Matter and Radiation

Energy of the incident photon on the metal surface is $3 W and then 5 W, where W$ is the work function for that metal. The ratio of velocities of emitted photoelectrons is

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