The threshold frequency a photosensitive metal is 3-3times-10-4-Hz- If light of frequency 8-2times-10-14-Hz is incident on this metal- The cut-off voltage the photoelectric emission is nearly2V3V5V1V

Question

Toppr Admin · Accepted Answer

The correct option is A 2VGiven -xA0-xA0- -x3BD-th-3-3-xD7-1014-xA0-Hz-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0- -x3BD-8-2-xD7-1014-xA0-HzCut-off voltage-xA0- eVo-h-x3BD-x2212-h-x3BD-thOr-xA0-xA0- Vo-he-x3BD-x2212-x3BD-th-x27F9-xA0-Vo-6-6-xD7-10-x2212-341-6-xD7-10-x2212-19-8-2-x2212-3-3-xD7-1014-2-xA0-V

The threshold frequency for a photosensitive metal is $3.3 \times 10^{4}$ Hz. If light of frequency $8.2 \times 10^{14}$ Hz is incident on this metal. The cut-off voltage for the photoelectric emission is nearly
2V
3V
5V
1V

The correct option is A 2V
Given : $ν_{t h} = 3.3 \times 10^{14} H z$ $ν = 8.2 \times 10^{14} H z$
Cut-off voltage $e V_{o} = h ν - h ν_{t h}$
Or $V_{o} = \frac{h}{e} (ν - ν_{t h})$
$⟹ V_{o} = \frac{6.6 \times 10^{- 34}}{1.6 \times 10^{- 19}} (8.2 - 3.3) \times 10^{14} = 2 V$

The threshold frequency for a photosensitive metal is 3.3×104Hz. If light of frequency 8.2×1014Hz is incident on this metal. The cut-off voltage for the photoelectric emission is nearly2V3V5V1V

The correct option is A 2VGiven : νth=3.3×1014 Hz ν=8.2×1014 HzCut-off voltage eVo=hν−hνthOr Vo=he(ν−νth)⟹ Vo=6.6×10−341.6×10−19(8.2−3.3)×1014=2 V

The threshold frequency for a photosensitive metal is $3.3 \times 10^{4}$ Hz. If light of frequency $8.2 \times 10^{14}$ Hz is incident on this metal. The cut-off voltage for the photoelectric emission is nearly
2V
3V
5V
1V

The correct option is A 2V
Given : $ν_{t h} = 3.3 \times 10^{14} H z$ $ν = 8.2 \times 10^{14} H z$
Cut-off voltage $e V_{o} = h ν - h ν_{t h}$
Or $V_{o} = \frac{h}{e} (ν - ν_{t h})$
$⟹ V_{o} = \frac{6.6 \times 10^{- 34}}{1.6 \times 10^{- 19}} (8.2 - 3.3) \times 10^{14} = 2 V$