For a certain metal, the threshold frequency is v0. If light of frequency 2v0 is incident on it the electron comes out with a maximum velocity of 4 × 106 m/s. If light of frequency of 5v0 is incident on it, the maximum velocity of the photo electron will be

In first case, 12mv12=h2v0−hv0=hv0….(1) and In second case, 12mv22=h5v0−hv0=4hv0. From⁡(1) and (2), 12mv22=4×12mv12 or v2=2v1=2×4×106=8×106m/s

For a certain metal, the threshold frequency is v0. If light of frequency 2v0 is incident on it the electron comes out with a maximum velocity of 4 × 106 m/s. If light of frequency of 5v0 is incident on it, the maximum velocity of the photo electron will be

In first case, 12mv12=h2v0−hv0=hv0….(1) and In second case, 12mv22=h5v0−hv0=4hv0. From⁡(1) and (2), 12mv22=4×12mv12 or v2=2v1=2×4×106=8×106m/s

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For a certain metal, the threshold frequency is v₀. If light of frequency 2v₀ is incident on it the electron comes out with a maximum velocity of 4 × 10⁶ m/s. If light of frequency of 5v₀ is incident on it, the maximum velocity of the photo electron will be

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12 × 10⁶ m/s

8 × 10⁶ m/s

16 × 10⁶ m/s

2 × 10⁶ m/s

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Detailed Solution

$In first case, \frac{1}{2} m v_{1}^{2} = h (2 v_{0}) - h v_{0} = h v_{0} \dots . (1) and$

$In second case, \frac{1}{2} m v_{2}^{2} = h (5 v_{0}) - h v_{0} = 4 h v_{0} .$

$From (1) and (2), \frac{1}{2} {mv}_{2}^{2} = 4 \times \frac{1}{2} {mv}_{1}^{2}$

$or v_{2} = 2 v_{1} = 2 \times (4 \times 10^{6}) = 8 \times 10^{6} m / s$

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For a certain metal, the threshold frequency is v0. If light of frequency 2v0 is incident on it the electron comes out with a maximum velocity of 4 × 106 m/s. If light of frequency of 5v0 is incident on it, the maximum velocity of the photo electron will be

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For a certain metal, the threshold frequency is v₀. If light of frequency 2v₀ is incident on it the electron comes out with a maximum velocity of 4 × 10⁶ m/s. If light of frequency of 5v₀ is incident on it, the maximum velocity of the photo electron will be