-
The photoelectric threshold wavelength of silver is 3250 × 10–10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10–10 m is
(Given h = 4.14 × 10–15 eVs and c = 3 × 108 ms–1)
-
- ≈0.6 × 106 ms–1
- ≈61 × 103 ms–1
- ≈0.3 × 106 ms–1
- ≈6 × 105 ms–1
- A and D Both
Correct Option: E
Both answers are correct Given,
λ0 = 3250 × 10–10 m
λ = 2536 × 10–10 m
| φ = | = | = 3.82eV | ||
| λ0 | 3250 × 10-10 |
| hv = | = | = 4.89eV | ||
| λ | 2536 × 10-10 |
According to Einstein's photoelectric equation,
Kmax = hv – φ
KEmax = (4.89–3.82)eV=1.077 eV
| = 1.077 × 1.6 × 10–19 | 2 |
| ⇒ v = |  | |
| 9.1 × 10-31 |
or, v = 0.6 × 106 m/s or 6 × 105 m/s
