JEE Main 20146 April
- Exam
- JEE Main
- Year
- 2014
- Date
- 6 April
- Subject
- Physics
- Chapter
- Atomic Physics
JEE Main 2014 · 6 April · Physics · Atomic Physics
The radiation corresponding to 3 → 2 transition of hydrogen atom...
The radiation corresponding to transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of . If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to :
Options
- A1.8 eV
- B1.1 eV
- C0.8 eV
- D1.6 eV
Correct answer
B. 1.1 eV
Step-by-step solution
We know that
\text{R} = \dfrac{\text{m} υ}{\text{Bq}}
∴ υ = \dfrac{\text{Bqr}}{\text{m}}
∴ \text{KE} = \dfrac{1}{2} \text{m} {\text{v}}^{2}
= 0.79 eV
Now from photoelectric equation
∴ ϕ = \text{E} - \text{KE}
\text{R} = \dfrac{\text{m} υ}{\text{Bq}}
∴ υ = \dfrac{\text{Bqr}}{\text{m}}
∴ \text{KE} = \dfrac{1}{2} \text{m} {\text{v}}^{2}
= 0.79 eV
Now from photoelectric equation
∴ ϕ = \text{E} - \text{KE}