Class 11th

This is a Long Answer Type Questions as classified in NCERT Exemplar

ANS-When the beam of light gets exposed to the metal surface then the electrons get emitted from the metal. This effect is said to be photoelectric effect and the emitted electrons are said to be photoelectrons. The result of photoelectric experiment are as follows: -

(i) There is no time gap between the striking of a light beam and the ejection of electrons from the metal surface i.e., electrons emit as soon as a beam of light strikes the metal surface.

(ii) The number of electrons ejected is directly proportional to the intensity of light.

(iii) For every metal there is minimum emitting frequency which is said to be threshold frequency, ν _o .The kinetic energies of these electrons is directly proportional to the frequency of the light used. As per the quantum theory of electromagnetic radiation when a photon with energy greater than the threshold energy of the metal strikes the metal then the electrons get emitted from the metal without any delay and its kinetic energy depends upon the frequency of the light.

As per the quantum theory of electromagnetic radiation when a photon with energy greater than the threshold energy of the metal strikes the metal then the electrons get emitted from the metal without any delay and its kinetic energy depends upon the frequency of the light

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

U_i + K_i = U_f + K_f

0+1/2mv²= mgh + 1/2mv'

$\frac{{\left(v\text{'}\right)}^2}{2}=\frac{v^2}{2}=-gh$

(v')²=v²-2gh = v^'= $\sqrt[]{v^2-2gh}$ ……….1

Let speed after emerging be v₁ then

=1/2mv₁²=1/2[1/2mv'²]

1/2m(v₁)²=1/4m(v')²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^2}{2}-gh}$ ………….2

From eqn 1 and 2

$\frac{v\text{'}}{v_1}=\frac{\sqrt[]{v^2-2gh}}{\sqrt[]{v^2-2gh/\sqrt[]{2}}}=\sqrt[]{2}$

So v₁ = v'/ $\sqrt[]{2}$ =v²(v'/2)

v₁>v'/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v'

(d) as the velocity of the bullet changes to v' which is less than v' hence , path, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.