The threshold frequency for certain metal is 3.3× 1014Hz . If light of frequency 8.2× 1014 Hz is incident on the metal, the cut-off voltage of the photoelectric current will be:

Question

The threshold frequency for certain metal is 3.3× 1014Hz . If light of frequency 8.2× 1014 Hz is incident on the metal, the cut-off voltage of the photoelectric current will be: (A) 4.9 V (B) 3.0 V (C) 2.0 V (D) 1.0 V

Tardigrade Admin · Accepted Answer

In photoelectric effect energy is conserved.
In photoelectric effect Einstein's equation is given by.
Photon energy

= K E

of electron + work function i.e.,

h v = e V_{s} + h v_{0}

where

V_{s}

is stopping potential and

v_{0}

is the threshold frequency.

∴

V_{s} = \frac{h}{e} (v - v_{0})

...(i)
Given,

v_{0} = 3.3 \times 10^{14} Hz, v = 8.2 \times 10^{14} Hz,

h = 6.6 \times 10^{- 34} J - s .

e = 1.6 \times 10^{- 19} C

Substituting the values in Eq. (i), we get

V_{s} = \frac{6.6 \times 10 ^{- 34} \times ( 8.2 \times 10 ^{14} - 3.3 \times 10 ^{14} )}{1.6 \times 10 ^{- 19}}

= \frac{6.6 \times 4.9}{1.6} \times 10^{- 1}

= 2.0 V

Q. The threshold frequency for certain metal is 3.3×1014Hz . If light of frequency 8.2×1014 Hz is incident on the metal, the cut-off voltage of the photoelectric current will be:

4.9 V

3.0 V

2.0 V

1.0 V

Q. The threshold frequency for certain metal is $3.3 \times 10^{14} Hz$ . If light of frequency $8.2 \times 10^{14}$ Hz is incident on the metal, the cut-off voltage of the photoelectric current will be: