The photon emitted during the
de-excitation from the first
excited level to the ground
state of hydoogen atom is
used to irradiate a photocathode
of a photocell, in which stopping potential of 5V is used . Calculate the work function of the cathode used
Answers
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