U.V. Light of wavelength 300 $A^{0}$ is incident on sodium surface. Then the maximum velocity of photo-electrons emitted is [neglect the work function of sodium].

3.8 \times 10^{6} m s^{- 1}

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4.8 \times 10^{6} m s^{- 1}

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6 \times 10^{5} m s^{- 1}

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3 \times 10^{5} m s^{- 1}

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Solution

The correct option is A $3.8 \times 10^{6} m s^{- 1}$
$φ = 0$
$K_{m a x} = \frac{h c}{λ} - φ$
$\frac{1}{2} m v^{2} = \frac{20 \times 10^{- 26}}{300 \times 10^{- 10}} - 0$
$v = 3.8 \times 10^{6} m s^{- 1}$
So, the answer is option (A).

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U.V. Light of wavelength 300 A0 is incident on sodium surface. Then the maximum velocity of photo-electrons emitted is [neglect the work function of sodium].

The correct option is A 3.8×106ms−1φ=0 Kmax= hcλ−φ 12mv2 = 20×10−26300×10−10−0 v= 3.8×106 ms−1So, the answer is option (A).

U.V. Light of wavelength 300 $A^{0}$ is incident on sodium surface. Then the maximum velocity of photo-electrons emitted is [neglect the work function of sodium].

The correct option is A $3.8 \times 10^{6} m s^{- 1}$
$φ = 0$
$K_{m a x} = \frac{h c}{λ} - φ$
$\frac{1}{2} m v^{2} = \frac{20 \times 10^{- 26}}{300 \times 10^{- 10}} - 0$
$v = 3.8 \times 10^{6} m s^{- 1}$
So, the answer is option (A).