A 5 W source emits monochromatic light of wavelength 5000 mathringA . When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m , the number of photoelectrons liberated will be reduced by a factor of

Question

A 5 W source emits monochromatic light of wavelength 5000 mathringA . When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m , the number of photoelectrons liberated will be reduced by a factor of (A)  4  (B)  8  (C)  16  (D)  2

Tardigrade Admin · Accepted Answer

Intensity of light is inversely proportional to square of distance.
ie,

I \propto \frac{1}{r ^{2}}

or

\frac{I _{2}}{I _{1}} = \frac{( r _{1} ) ^{2}}{( r _{2} ) ^{2}}

Given,

r_{1} = 0.5 m

,

r_{2} = 1.0 m

Therefore,

\frac{I _{2}}{I _{1}} = \frac{( 0.5 ) ^{2}}{( 1 ) ^{2}}

= \frac{1}{4}

Now, since number of photoelectrons emitted per second is directly proportional to intensity, so number of electrons emitted would decrease by factor of

4

.

Q. A $5 W$ source emits monochromatic light of wavelength $5000 \overset{˚}{A}$ . When placed $0.5 m$ away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of $1.0 m$ , the number of photoelectrons liberated will be reduced by a factor of

$4$

$8$

$16$

$2$

Q. A 5W source emits monochromatic light of wavelength 5000A˚ . When placed 0.5m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0m , the number of photoelectrons liberated will be reduced by a factor of

4

8

16

2

Q. A $5 W$ source emits monochromatic light of wavelength $5000 \overset{˚}{A}$ . When placed $0.5 m$ away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of $1.0 m$ , the number of photoelectrons liberated will be reduced by a factor of

$4$

$8$

$16$

$2$