The threshold frequency for certain metal is 3.3× 1014 Hz . If the 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× 1014 Hz . If the light of frequency 8.2× 1014 Hz is incident on the metal, the cut-off voltage of the photoelectric current will be (A) 4.9V (B) 3.0 V (C) 2.0 V (D) 1V

Tardigrade Admin · Accepted Answer

From relation eVs=h(v - v0) or Vs= threshold or cut off voltage =(h/e)(v - v0) = (6.6 × (10)- 34/1.6 × (10)- 19) (.8.2-3.3.)× (10)14 = (6 .6 × 4 .9 × 10- 1/1 .6) = 2V

Q. The threshold frequency for certain metal is $3.3 \times 1 0^{14}$ $Hz$ . If the light of frequency $8.2 \times 1 0^{14}$ $Hz$ is incident on the metal, the cut-off voltage of the photoelectric current will be

$4.9 V$

$3.0 V$

$2.0 V$

$1 V$

From relation
$e V_{s} = h (v - v_{0})$
or $V_{s} =$ threshold or cut off voltage
$= \frac{h}{e} (v - v_{0})$
$= \frac{6.6 \times ( 10 ) ^{- 34}}{1.6 \times ( 10 ) ^{- 19}} (8.2 - 3.3) \times (10)^{14}$
$= \frac{6.6 \times 4.9 \times 1 0 ^{- 1}}{1.6} = 2 V$

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

4.9V

3.0V

2.0V

1V