1 Marks Question

Question 11 Mark

In an experiment on photoelectric effect, a photon is incident on an electron from one direction and the photoelectron is emitted almost in the opposite direction. Does this violate conservation of momentu?

Answer

No, it does not violate the principle of conservation of momentum. In the photon-electron collision, the energy and momentum are conserved.

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Question 21 Mark

Can a photon be deflected by an electric field? By a magnetic field?

Answer

Photons are electrically neutral. Hence, they are not deflected by electric and magnetic fields.

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Question 31 Mark

The threshold wavelength of a metal is $\lambda_0.$ Light of wavelength slightly less $\tan \lambda_0.$ is incident on an insulated plate made of this metal. It is found that photoelectrons are emitted for some time and after that the emission stops. Explain.

Answer

When light of wavelength less than $\lambda_0.$ is incident on the metal surface, the free electrons of the metal will gain energy and come out of the metal surface. As the metal plate is insulated (it is not connected with the battery), the free electrons of the metal will not be replaced by the other electrons. Hence, photoelectron emission will stop after some time.

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Question 41 Mark

Consider the situation described in the previous problem. Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.

Answer

For a perfectly reflecting solid sphere of radius ‘r’ kept in the path of a parallel beam of light of large aperture with intensity ‘I’, force exerted $=\frac{\pi\text{r}^2\text{I}}{\text{C}}$

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Question 51 Mark

A sphere of radius 1.00cm is placed in the path of a parallel beam of light of large aperture. The intensity of the light is 0.5W/cm^-2. If the sphere completely absorbs the radiation falling on it, find the force exerted by the light beam on the sphere.

Answer

We know,

If a perfectly reflecting solid sphere of radius ‘r’ is kept in the path of a parallel beam of light of large aperture if intensity is I,

$\text{Force}=\frac{\pi\text{r}^2\text{I}}{\text{C}}$

$\text{I}=0.5\text{W/m}^2.\text{r}=1\text{cm},\text{C}=3\times10^8\text{m/s}$

$\text{Force}=\frac{\pi(1)^2\times0.5}{3\times10^8}=\frac{3.14\times0.5}{3\times10^8}$

$=0.523\times10^{-8}=5.2\times10^{-9}\text{N.}$

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