Atomic Structure - Result Question 94
####70. The work function $(\phi)$ of some metals is listed below. The number of metals which will show photoelectric effect when light of $300 nm$ wavelength falls on the metal is
(2011)
| Metal | $Li$ | $Na$ | $K$ | $Mg$ | $Cu$ | $Ag$ | $Fe$ | $Pt$ | $W$ |
|---|---|---|---|---|---|---|---|---|---|
| $\Phi(eV)$ | 2.4 | 2.3 | 2.2 | 3.7 | 4.8 | 4.3 | 4.7 | 6.3 | 4.75 |
Show Answer
Answer:
Correct Answer: 70. (9)
Solution:
- Energy of photon
$$ =\frac{h c}{\lambda} J=\frac{h c}{e \lambda} eV=\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9} \times 1.602 \times 10^{-19}}=4.14 eV $$
For photoelectric effect to occur, energy of incident photons must be greater than work function of metal. Hence, only Li, Na, $K$ and $Mg$ have work functions less than $4.14 V$.
