The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Find the wavelength of the radiation in $A 0$

Question

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Answer 1

From the Einstein's photo electric equation, we get
$\frac{h c}{λ} = e V_{0} + ϕ$
$∴ \frac{h c}{λ} = 10.4 e V + 1.7 e V$
$∴ \frac{h c}{λ} = 12.1 e V$
$∴ λ = \frac{h c}{1 2 . 1 \times 1 . 6 \times 1 0 ^{- 19}}$
$= \frac{6 . 6 3 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{1 2 . 1 \times 1 . 6 \times 1 0 ^{- 19}}$
$= 1026 A °$
Now, the transition between $3 \to 1$
we get
$\frac{1}{λ _{1}} = R (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = R (1 - \frac{1}{9})$
$= \frac{8}{9} R = \frac{8 \times 1 . 0 9 7 \times 1 0 ^{7}}{9}$
$∴ λ = 1026 A °$
$∴$ The energy level in hydrogen atom which will emit this wavelength is $n = 3$ to $n = 1$

Answer 2

From the Einstein's photo electric equation, we get

\frac{h c}{λ} = e V_{0} + ϕ

∴ \frac{h c}{λ} = 10.4 e V + 1.7 e V

∴ \frac{h c}{λ} = 12.1 e V

∴ λ = \frac{h c}{1 2 . 1 \times 1 . 6 \times 1 0 ^{- 19}}

= \frac{6 . 6 3 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{1 2 . 1 \times 1 . 6 \times 1 0 ^{- 19}}

= 1026 A °

Now, the transition between

3 \to 1

we get

\frac{1}{λ _{1}} = R (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = R (1 - \frac{1}{9})

= \frac{8}{9} R = \frac{8 \times 1 . 0 9 7 \times 1 0 ^{7}}{9}

∴ λ = 1026 A °

∴

The energy level in hydrogen atom which will emit this wavelength is

n = 3

to

n = 1