Question

The minimum energy required for the emission of photoelectron from the surface of a metal is $ \, 4.95 \, \times \, 10^{- 19} \, J$ . Calculate the critical frequency of the photon required to eject the electron. $h= \, 6.6 \, \times 10^{- 34} \, J \, sec$ .

• A. $7.5 \, \times 10^{14}s^{- 1}$
• B. $7.5 \, \times 10^{13}s^{- 1}$
• C. $7.5 \, \times 10^{16}s^{- 1}$
• D. $7.5 \, \times 10^{19}s^{- 1}$

Answer 1

Answer: (A)

Minimum energy required for the emission of photoelectrons $=4.95 \, \times 10^{- 19}J$
Let the critical frequency or threshold frequency of the photon to eject the electron is $ \, v \, s e c^{- 1}$ .
We know, $E \, = \, h\upsilon \, \, or \, \upsilon \, = \, \frac{E}{h} \, = \, \frac{4.95 \times 10^{- 19}}{6.6 \times 10^{- 34}} \, = \, 7.5 \, \times 10^{14}s^{- 1}$