In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength () of in

Question

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength ( $λ$ ) of incident light and the corresponding stopping potential ( $V_{0}$ ) are given below:

$λ (μ m)$	$V_{0} (Volt)$
0.3	2.0
0.4	1.0
0.5	0.4

Given that $c = 3 \times 10^{8} {ms}^{- 1}$ and $e = 1.6 \times 10^{- 19} C,$ Planck's constant (in units of J s) found from such an experiment is [2016]

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(2)

$\frac{h c}{e λ_{1}} - \frac{ϕ}{e} = V_{0_{1}}$ and $\frac{h c}{e λ_{2}} - \frac{ϕ}{e} = V_{0_{2}}$ $(∵ \frac{h c}{λ} - ϕ = e V_{0})$

$∴$ $e = 1.6 \times 10^{- 19} C,$

$∴$ $h = \frac{e (V_{0_{1}} - V_{0_{2}}) λ_{1} λ_{2}}{(λ_{2} - λ_{1}) c}$

From the first two values given in data,

$h = \frac{1.6 \times 10^{- 19} (2 - 1) \times 0.4 \times 0.3 \times 10^{- 6}}{0.1 \times 3 \times 10^{8}}$

$h = 0.64 \times 10^{- 33} = 6.4 \times 10^{- 34} J-s$

Similarly, if we calculate $h$ for the last two values of data, we get the same value of $h = 6.4 \times 10^{- 34} J-s$

Solution

1 $6.0 \times 10^{- 34}$

2 $6.4 \times 10^{- 34}$

3 $6.6 \times 10^{- 34}$

4 $6.8 \times 10^{- 34}$

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Solution

1 6.0×10-34

2 6.4×10-34

3 6.6×10-34

4 6.8×10-34

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1 $6.0 \times 10^{- 34}$

2 $6.4 \times 10^{- 34}$

3 $6.6 \times 10^{- 34}$

4 $6.8 \times 10^{- 34}$