A photoelectric cell is illuminated by a poin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$.Since the number of photoelectrons emitted

Vedclass · Accepted Answer

Number of electrons emitted is a quarter of the initial number

Vedclass · Answer

(C) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$.Since the number of photoelectrons emitted per second is directly proportional to the intensity of incident light,we have $n \propto I \propto \frac{1}{d^2}$.Given initial distance $d_1 = 1\;m$ and final distance $d_2 = 2\;m$.Therefore,the ratio of the number of electrons is $\frac{n_2}{n_1} = \left( \frac{d_1}{d_2} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Thus,$n_2 = \frac{n_1}{4}$.The energy of each emitted electron depends on the frequency of the incident light,not on the intensity,so the energy remains unchanged.

$A$ photoelectric cell is illuminated by a point source of light $1\;m$ away. When the source is shifted to $2\;m$,then:

Similar Questions

In which of the following pairs of categories is it more difficult to determine the relationship to other taxa at the same level, thus making the problem of classification more complex?

The yellow colour solution of $Na_2CrO_4$ changes to orange-red on passing $CO_2$ gas due to the formation of

For all values of $a$ and $b$,the line $(a+2b)x + (a-b)y + (a+5b) = 0$ passes through a fixed point. Find that point.

If the force is given by $F = at + bt^2$,where $t$ is time,the dimensions of $a$ and $b$ are:

The probability of a student qualifying for $IITJEE$ and $EAMCET$ are $\frac{1}{5}$ and $\frac{3}{5}$ respectively. The probability that the student qualifies for at least one of these tests is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module