The radiation corresponding to 3->2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5 × 10-4 T. Assume that the radius of the largest circular path followed by these electrons is 7mm, the work function of the metal is: (Mass of electron = 9.1 × 10-31 kg)

h ν = 3 . 4 − 1 . 5 1 = 1 . 8 9 e V As we know that radius of circular path in magnetic field is given as r = 2 m K q B ⇒ K = r 2 q 2 B 2 2 m  ⇒ K = ( 7 × 1 0 − 3 ) 2 × ( 1 . 6 × 1 0 − 1 9 ) × ( 5 × 1 0 − 4 ) 2 2 × 9 . 1 × 1 0 − 3 1 e V = 1 0 7 . 7 × 1 0 − 2 e V  = 1.08 eV ⇒ ? = h ν − K = 1 . 8 9 − 1 . 0 8 = 0 . 8 1 e V

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Physics Class 12th Dual Nature of Radiation and Matter

The radiation corresponding to 3->2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5 × 10^-4 T. Assume that the radius of the largest circular path followed by these electrons is 7mm, the work function of the metal is: (Mass of electron = 9.1 × 10^-31 kg)

Option 1 -

0.82 eV

Option 2 -

1.88 eV

Option 3 -

1.36 eV

Option 4 -

0.16 eV

3 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$h ν = 3.4 - 1.51 = 1.89 e V$

As we know that radius of circular path in magnetic field is given as

$r = \frac{\sqrt[]{2 m K}}{q B} \Rightarrow K = \frac{r^{2} q^{2} B^{2}}{2 m}$

$\Rightarrow K = \frac{{(7 \times 1 0^{- 3})}^{2} \times (1.6 \times 1 0^{- 19}) \times {(5 \times 1 0^{- 4})}^{2}}{2 \times 9.1 \times 1 0^{- 31}} e V = 107.7 \times 1 0^{- 2} e V$ = 1.08 eV

$\Rightarrow ? = h ν - K = 1.89 - 1.08 = 0.81 e V$

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