NEET Physics Photo Electric Effect, X- Rays & Matter Waves NEET PYQ-Photo Electric Effect,X-Rays

                        A light source is at a distance d from a photoelectric cell, then the number of photoelectrons emitted from the cell is n. If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become:                                                                                                                                                                    [AIPMT 2001]

A) \[\frac{n}{2}\]

B)      2n

C) 4 n

D)                  n

Correct Answer: C

Solution :

Intensity of Sight source is given by \[I\,\propto \,\,\frac{1}{{{d}^{2}}}\] where d is the distance of tight source from the cell.

or \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{d}_{2}}}{{{d}_{1}}} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{4}\]

or \[{{I}_{2}}=4{{I}_{1}}\]

As number of photoelectrons emitted is directly proportional to intensity, so number of photoelectrons emitted will become 4 times, i.e., 4n.