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

Question

The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz. is incident on this metal, the cut-off voltage for the photoelectron emission is nearly (A) 1 V (B) 2 V (C) 3 V (D) 5 V

Tardigrade Admin · Accepted Answer

V 0=( E Ph - W / e )=( h ( v - v 0)/ e ) =(6.62 × 10-34(8.2 × 1014-3.3 × 1014)/1.6 × 10-19) =(6.62 × 10-34/1.6) × 4.9 × 1014+19 =(6.62 × 4.9/1.6) × 10-1=2 volt

Q. The threshold frequency for a photosensitive metal is $3.3 \times 1 0^{14} Hz .$ If light of frequency $8.2 \times 1 0^{14} Hz .$ is incident on this metal, the cut-off voltage for the photoelectron emission is nearly

1 V

2 V

3 V

5 V

$V_{0} = \frac{E _{P h} - W}{e} = \frac{h ( v - v _{0} )}{e}$
$= \frac{6.62 \times 1 0 ^{- 34} ( 8.2 \times 1 0 ^{14} - 3.3 \times 1 0 ^{14} )}{1.6 \times 1 0 ^{- 19}}$
$= \frac{6.62 \times 1 0 ^{- 34}}{1.6} \times 4.9 \times 1 0^{14 + 19}$
$= \frac{6.62 \times 4.9}{1.6} \times 1 0^{- 1} = 2$ volt

Q. The threshold frequency for a photosensitive metal is 3.3×1014Hz. If light of frequency 8.2×1014Hz. is incident on this metal, the cut-off voltage for the photoelectron emission is nearly

1 V

2 V

3 V

5 V

V0​=eEPh​−W​=eh(v−v0​)​ =1.6×10−196.62×10−34(8.2×1014−3.3×1014)​ =1.66.62×10−34​×4.9×1014+19 =1.66.62×4.9​×10−1=2 volt

Q. The threshold frequency for a photosensitive metal is $3.3 \times 1 0^{14} Hz .$ If light of frequency $8.2 \times 1 0^{14} Hz .$ is incident on this metal, the cut-off voltage for the photoelectron emission is nearly

$V_{0} = \frac{E _{P h} - W}{e} = \frac{h ( v - v _{0} )}{e}$
$= \frac{6.62 \times 1 0 ^{- 34} ( 8.2 \times 1 0 ^{14} - 3.3 \times 1 0 ^{14} )}{1.6 \times 1 0 ^{- 19}}$
$= \frac{6.62 \times 1 0 ^{- 34}}{1.6} \times 4.9 \times 1 0^{14 + 19}$
$= \frac{6.62 \times 4.9}{1.6} \times 1 0^{- 1} = 2$ volt