JEE Main 12 Jan 2019 Evening Question 10
Question: In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 220 nm
B) 1700 nm
C) 250 nm
D) 2020 nm
Show Answer
Answer:
Correct Answer: C
Solution:
The minimum wavelength of emitted photons is $ \frac{hc}{\Delta E}=\frac{1240}{5.6-0.7}\simeq 250nm $
