Question

A photon emitted during the de – excitation of electron from a sate n to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2eV, in a photo cell, with a stopping potential of 0.55 V. Obtain the value of the quantum number of the state n.

Answer 1

The value of quantum number of the state n=4

Explanation:

Let $n_1=n$

First excited state=$n_2=2$

Work function=$\phi=2e V$

Stopping potential=$0.55 V$

We know that

Energy of photon=K.E+work function

$E=eV_0+\phi$

Where $V_0=$ Stopping potential

$\phi=$ Work function

Substitute the values

$E=0.55e+2e V=2.55e V$

Energy , $E_n=-\frac{13.6 eV}{n^2}$

Using the formula

Energy of electron in ground state=$E_2=-\frac{13.6}{2^2}=-3.4 eV$

Energy of electron in excited state n=$E_n=-\frac{13.6e V}{n^2}$

$E=E_n-E_2=-\frac{13.6}{n^2}+3.4$

$2.55e V-3.4=-\frac{13.6}{n^2}$

$-0.85=\frac{-13.6}{n^2}$

$n^2=\frac{13.6}{0.85}$

$n=\sqrt{\frac{13.6}{0.85}}=4$

Hence, the value of quantum number of the state n=4

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Answer 2

Answer:

Let n_1=nn

1

=n

First excited state=n_2=2n

2

=2

Work function=\phi=2e Vϕ=2eV

Stopping potential=0.55 V0.55V

We know that

Energy of photon=K.E+work function

E=eV_0+\phiE=eV

0

+ϕ

Where V_0=V

0

= Stopping potential

\phi=ϕ= Work function

Substitute the values

E=0.55e+2e V=2.55e VE=0.55e+2eV=2.55eV

Energy , E_n=-\frac{13.6 eV}{n^2}E

n

=−

n

2

13.6eV

Using the formula

Energy of electron in ground state=E_2=-\frac{13.6}{2^2}=-3.4 eVE

2

=−

2

2

13.6

=−3.4eV

Energy of electron in excited state n=E_n=-\frac{13.6e V}{n^2}E

n

=−

n

2

13.6eV

E=E_n-E_2=-\frac{13.6}{n^2}+3.4E=E

n

−E

2

=−

n

2

13.6

+3.4

2.55e V-3.4=-\frac{13.6}{n^2}2.55eV−3.4=−

n

2

13.6

-0.85=\frac{-13.6}{n^2}−0.85=

n

2

−13.6

n^2=\frac{13.6}{0.85}n

2

=

0.85

13.6

n=\sqrt{\frac{13.6}{0.85}}=4n=

0.85

13.6

=4

Hence, the value of quantum number of the state n=4