JEE PYQ: Atomic Physics Question 28
Question 28 - 2020 (03 Sep Shift 2)
If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is:
Planck’s constant, $h = 6.63 \times 10^{-34}$ J.s. Speed of light, $c = 3 \times 10^8$ m/s.
(1) 1.1 eV
(2) 2.0 eV
(3) 1.5 eV
(4) 3.1 eV
Show Answer
Answer: (4)
Solution
Wavelength of photon, $\lambda = 400$ nm. A photodiode can detect a wavelength corresponding to the energy of band gap. Band gap $E_g = \frac{hc}{\lambda} = \frac{1237.5}{400} = 3.09$ eV $\approx 3.1$ eV.
BETA
