The minimum frequency of light source to get photocurrent, from a metal surface having work function 2eV is

3.2 \times 10^{14} H z

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1.6 \times 10^{14} H z

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4.3 \times 10^{14} H z

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4.8 \times 10^{14} H z

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Solution

The correct option is D $4.8 \times 10^{14} H z$
$K E ⩾ 0$
$h v - φ ⩾ 0$
$6.63 \times 10^{- 34} \times V - 2 \times 1.6 \times 10^{- 19} ⩾ 0$
$V ⩾ 4.8 \times 10^{14} H z$
So, the answer is option (D).

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Similar questions

Q. A light whose frequency is equal to

6 \times 10^{14} H z

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(h = 6.63 \times 10^{- 34} J s, 1 e V = 1.6 \times 10^{- 19} J)

. The maximum energy of electrons emitted will be:

Q. The frequency range of visible light is from

3.75 \times 10^{14} H z

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The threshold frequency for a certain metal is 3.3 × 10¹⁴Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Q. The threshold frequency for a photosensitive metal is

3.3 \times 10^{14} H z

. If light of frequency

8.2 \times 10^{14} H z

is incident on this metal, the cutoff voltage for the photoelectric emission is nearly :

Q. The threshold frequency for certain metal is

3.3 \times 10^{14} H z

. If light of

8.2 \times 10^{14} H z

frequency is incident on the metal, then the cut-off voltage of the photoelectric current will be :

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The minimum frequency of light source to get photocurrent, from a metal surface having work function 2eV is

The correct option is D 4.8×1014HzKE ⩾0 hv−φ⩾0 6.63×10−34×V −2×1.6×10−19⩾0 V ⩾ 4.8×1014 HzSo, the answer is option (D).

The correct option is D $4.8 \times 10^{14} H z$
$K E ⩾ 0$
$h v - φ ⩾ 0$
$6.63 \times 10^{- 34} \times V - 2 \times 1.6 \times 10^{- 19} ⩾ 0$
$V ⩾ 4.8 \times 10^{14} H z$
So, the answer is option (D).