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Stopping Potential Revisited

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Question

Light of wavelength $4000 \circ A$ falls on a photosensitive metal and a negative 2V potential stops the emitted electrons. The work function of the material (in eV) is approximately $(h = 6.6 \times 10^{- 34} J s, e = 1.6 \times 10^{- 19} C, c = 3 \times 10^{8} m s^{- 1})$

1.1

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2.2

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3.1

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Solution

The correct option is A 1.1
Energy of incident light $E (e V) = \frac{12375}{4000} = 3.09 e V$
Stopping potential is-2V so $K_{m a x}; = 2 e V$
Hence by using $E = W_{0} + K_{m a x}; W = 1.09 e V \approx 1.1 e V$

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4.36 \times 10^{- 7}

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4.2 e V

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= 3 \times 10^{8} m / s

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= 1.6 \times 10^{- 19} C

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2200 A^{o}

4.1

4000 \circ A

(h = 6.6 \times 10^{- 34} J s, e = 1.6 \times 10^{- 19} C, c = 3 \times 10^{8} m s^{- 1})

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