Photons emitted when electrons in a H-atom make transition from a higher energy state to lower energy state- whose difference in angular momentum is h - - are made to be incident on a sample of sodium metal work function- W -2-3 eV - What will be the kinetic energy of the photoelectrons- A- 7-9 eVB- 0-25 eVC- 10-45 eVD- 9-79 eV

The correct options are B 0.25 eV D 9.79 eVdifference in angular momentum = nbsp;hπ = (n_2 n_1)h2π ∴(n_2 n_1)=2 only two transitions are possible for which ...

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Standard XII

Physics

Characteristics of Bohr's Atomic Model

Photons emitt...

Question

Photons emitted when electrons in a H-atom make transition from a higher energy state to lower energy state, whose difference in angular momentum is $\frac{h}{π}$ , are made to be incident on a sample of sodium metal (work function, $W = 2.3 e V$ ). What will be the kinetic energy of the photoelectrons?

7.9 e V

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0.25 e V

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10.45 e V

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9.79 e V

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Solution

The correct options are
B $0.25 e V$
D $9.79 e V$
difference in angular momentum = $\frac{h}{π}$ = $(n_{2} - n_{1}) \frac{h}{2 π}$
$∴ (n_{2} - n_{1}) = 2$
only two transitions are possible for which difference in shell energy is more than $2.3 e V$ which are:
$E_{3 \to 1} = 12.09 e V$ and $E_{4 \to 2} = 2.55 e V$
and $K . E = h ν - ϕ$ where, $ϕ$ = work function
so possible K.E. of photoelectrons are $12.09 - 2.3 = 9.79 e V$ and $2.55 - 2.3 = 025 e V$

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Photons emitted when electrons in a H-atom make transition from a higher energy state to lower energy state, whose difference in angular momentum is hπ , are made to be incident on a sample of sodium metal (work function, W=2.3 eV). What will be the kinetic energy of the photoelectrons?