Question

4.
A doubly ionized lithium atom in an excited state (n = 6) emits a photon of energy 4.25 eV. What
are the quantum number (n) and the energy (E) of the final state?
(a) n = 2, E = -30.6 eV
(b) n = 3. E = -13.6 eV
(c) n = 4. E= -7.65 eV
(d) n=5, E = -4.90 eV​

Answer 1

Given

An electron is in sixth excited state and then it emits a photon of energy 4.25 eV.

To Find

• Quantum number (n) of the final state.

• Energy of the final state.

Concept

Energy is supplied to an electron to excite it into a higher orbit. Conversely, photon is emitted when it de- excites into lower orbit.

Formula

Energy of electron in nth orbit = $- \frac{13.6\:Z^2}{n^2}$ eV.

Radius of nth orbit = $\frac{0.529\: n^2}{Z}$ Ā

Velocity of electron in nth orbit = $\frac{2.18×10^6\:Z}{n}$ m/s

Calculations

Energy of electron of $Li^{2+}$ (Z = 3) in an orbit having n = 6

= - $- \frac{13.6\:Z^2}{n^2}$ eV.

= - $- \frac{13.6\:(3)^2}{(6)^2}$ eV.

= - $- \frac{13.6\:×\:9}{36}$ eV.

= - 3.4 eV

Now, Let energy of the orbit in which it will go be x.

Then , x + 4.25 = -3.4

=> x = - 7.65 eV.

$\boxed{\boxed{\pink{Hence,\:energy\:of\:electron\:in\:final\:state\:is\:-7.65\:eV}}}$

Now, using the same formula :- $E\:=\: - \frac{13.6\:Z^2}{n^2}$ eV , we can find n.

$\pink{ \rightarrow}$ We have to find the value of n for an orbit whose energy is -7.65 eV.

=> -7.65 = $- \frac{13.6\:Z^2}{n^2}$ eV.

=> 0.5625 = $\frac{9}{n^2}$

=> 0.0625 = $\frac{1}{n^2}$

=> $\frac{1}{0.0625} = n^2$

=> $\frac{1}{0.0625} = n^2$

=> $16 = n^2$

=> n = 4

$\boxed{\boxed{\pink{Hence,\:quantum\:number\:of\:final\:state\:is\:6}}}$

Therefore, the correct alternative is (c).