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Calculate the value of the Planck's constant $h$ if photoelectrons emitted from a surface of a certain metal by light of frequency $2.2 \times 10^{15} H z$ are fully retarded by a reverse potential of $6.6 V$ and those ejected by light of frequency $4.6 \times 10^{15} H z$ by a reverse potential of $16.5 V$ .

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$6.6 \times 10^{- 24} J s$

$3.6 \times 10^{- 34} J s$

$6.6 \times 10^{- 34} J s$

$6.6 \times 10^{- 14} J s$

answer is A.

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Detailed Solution

According to Einstein's photo electric equation, we have

$h v_{1} = ϕ_{0} + e V_{1}$

$h v_{2} = ϕ_{0} + e V_{2}$

Subtracting Equation (1) from Equation (2), we get

$h (v_{2} - v_{1}) = e (V_{2} - V_{1})$

$\Rightarrow h = \frac{(16.5 - 6.6) (1.6 \times 10^{- 19})}{(4.6 - 2.2) \times 10^{15}}$

$\Rightarrow h = 6.6 \times 10^{- 34} J s$

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