Question 80: Dual Nature of Matter and Radiation

Question

A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV )

Accepted Answer

Let's break down the problem step by step.
The photoelectric equation is given by:
$$K_{	ext{max}} = \frac{hc}{\lambda} - \phi$$
where:
$$K_{	ext{max}}$$ is the maximum kinetic energy of the electrons.
$$\frac{hc}{\lambda}$$ is the energy of the incident photon.
$$\phi$$ is the work function of the metal.
For the initial light source of wavelength $$\lambda$$, we know:
$$2 	ext{ eV} = \frac{hc}{\lambda} - 1 	ext{ eV}$$
Solving for $$\frac{hc}{\lambda}$$ gives:
$$\frac{hc}{\lambda} = 2 	ext{ eV} + 1 	ext{ eV} = 3 	ext{ eV}$$
Now, if the wavelength is halved to $$\frac{\lambda}{2}$$, the photon energy becomes:
$$\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = 2 	imes 3 	ext{ eV} = 6 	ext{ eV}$$
The maximum kinetic energy for the new wavelength is then:
$$K'_{	ext{max}} = 6 	ext{ eV} - 1 	ext{ eV} = 5 	ext{ eV}$$
Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated by a light source of wavelength $$\frac{\lambda}{2}$$ is 5 eV.
The correct answer is Option A.

Answer

5 eV

Answer

3 eV

Answer

2 eV

Answer

6 eV

Dual Nature of Matter and Radiation

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