Question 267: Dual Nature of Matter and Radiation

Question

If the electron in hydrogen atom jumps from third Bohr orbit to ground state directly and the difference between energies of the two states is radiated in the form of photons. If the work function of the material is 4.1 eV , then stopping potential is nearly
[Energy of electron in $n^{	ext {th }}$ orbit $=\frac{-13 \cdot 6}{n^2} \mathrm{eV}$ ]

Accepted Answer

Finding the energy difference when the electron jumps:
The electron moves from the 3rd orbit ($n=3$) to the ground state ($n=1$).
The energy in the $n^{	ext{th}}$ orbit is $E_n = \frac{-13.6}{n^2}$ eV.
So,

$E_1 = \frac{-13.6}{1^2} = -13.6$ eV

and

$E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} = -1.51$ eV.
The energy difference is:

$$E_{	ext{photon}} = E_1 - E_3 = (-13.6) - (-1.51) = -13.6 + 1.51 = -12.09 	ext{ eV}$$

But since energy emitted is positive, use $12.09$ eV.
Linking energy to the photon:
This energy is carried by the photon that is released:

$h
u = 12.09 	ext{ eV}$
Using the photoelectric equation to find stopping potential:
The photon causes electrons to be emitted from a material that has a work function ($\phi_0$) of $4.1$ eV.
The equation is:

$$h
u = \phi_0 + eV_s$$

where $V_s$ is the stopping potential.
Plug in the known values:

$12.09 = 4.1 + V_s$
Solve for $V_s$:

$V_s = 12.09 - 4.1 = 8.0 	ext{ V}$
So, the stopping potential is about $8 	ext{ V}$.

Answer

3 V

Answer

4 V

Answer

6 V

Answer

8 V

Dual Nature of Matter and Radiation

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