JEE Main 201912 JanuaryEvening Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 12 January
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Atomic Physics
JEE Main 2019 · 12 January · Evening Shift · Physics · Atomic Physics
In a Frank - Hertz experiment, an electron of energy 5.6eV passes...
In a Frank - Hertz experiment, an electron of energy passes through mercury vapour and emerges with an energy The minimum wavelength of photons emitted by mercury atoms is close to:
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
When electron pass through the mercury vapor, it losses some of its energy. The loss in of electron =\left(5 ⋅ 6 - 0.7\right)eV = 4.9eV∴= 4.9eV∴\lambda = \dfrac{1.24 \times {10}^{4}}{4.9}A \approx 250nm