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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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Similar questions

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})

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})

4000 A^{o}

h = 6.62 \times 10^{- 34} J s

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

2.0 e V

3300

A^{o}

h = 6.6 \times 10^{- 34} J s, 1 e V = 1.6 \times 10^{- 19} J

c = 3 \times 10^{8} m s^{- 1}

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

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