You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stopping Potential Revisited

Light of wave...

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $E = W_{0} + K_{m a x}; W = 1.09 e V \approx 1.1 e V$

Suggest Corrections

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 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}

2 \times 10^{- 7} m

2.5 V

e = 1.6 \times 10^{- 19} C,

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

1 A^{o}

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

e = 1.6 \times 10^{- 19} C

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

Join BYJU'S Learning Program