A copper ball of radius 1 cm and work function 4.47 eV is irradiated with ultraviolet radiation of wavelength 2500 mathringA. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is:

Question

A copper ball of radius 1 cm and work function 4.47 eV is irradiated with ultraviolet radiation of wavelength 2500 mathringA. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is: (A) 5.5×10-13C (B) 7.5×10-13 C (C) 4.5×10-12 C (D) 2.5×10-11C

Tardigrade Admin · Accepted Answer

From photo electric effect equation: ] h v=h v0+K . E max so Maximum kinetic energy will be putting the given values in the above equation K . E max = e × V so V will be V=(k × Q/r) :: q =5.5 × 10-13 C

$5.5 \times 1 0^{- 13} C$

$7.5 \times 1 0^{- 13} C$

$4.5 \times 1 0^{- 12} C$

$2.5 \times 1 0^{- 11} C$

From photo electric effect equation : ]
$h v = h v_{0} + K . E_{m a x}$
so Maximum kinetic energy will be
putting the given values in the above equation
$K . E_{m a x} = e \times V$
so $V$ will be
$V = \frac{k \times Q}{r}$
$:: q = 5.5 \times 1 0^{- 13} C$

5.5×10−13C

7.5×10−13C

4.5×10−12C

2.5×10−11C

From photo electric effect equation : ] hv=hv0​+K.Emax​ so Maximum kinetic energy will be putting the given values in the above equation K.Emax​=e×V so V will be V=rk×Q​ ::q=5.5×10−13C

$5.5 \times 1 0^{- 13} C$

$7.5 \times 1 0^{- 13} C$

$4.5 \times 1 0^{- 12} C$

$2.5 \times 1 0^{- 11} C$

From photo electric effect equation : ]
$h v = h v_{0} + K . E_{m a x}$
so Maximum kinetic energy will be
putting the given values in the above equation
$K . E_{m a x} = e \times V$
so $V$ will be
$V = \frac{k \times Q}{r}$
$:: q = 5.5 \times 1 0^{- 13} C$