During the transition of electron from state A to state C of a Bohr aom, the wavelength of emitted radiation is 2000 and it becomes 6000 when the electron jumps from state B to stacte C. Then the wavelength of the radiation emitted during the transition

Question

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 $\overset{°}{A}$ and it becomes 6000 $\overset{°}{A}$ when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is [2025]

Smart Achievers · Accepted Answer

(1)

For A to C : $\overset{°}{A}$ ... (i)

and for B to C : $E_{B} - E_{C} = \frac{h c}{6000 \overset{°}{A}}$ ... (ii)

Now for A to B : $E_{A} - E_{B} = (E_{A} - E_{C}) - (E_{B} - E_{C})$

$\frac{h c}{λ_{A B}} = \frac{h c}{2000} - \frac{h c}{6000}$

$\frac{1}{λ_{A B}} = \frac{1}{3000 \overset{°}{A}} \Rightarrow λ_{A B} = 3000 \overset{°}{A}$

Solution

1 3000 $\overset{°}{A}$

2 6000 $\overset{°}{A}$

3 4000 $\overset{°}{A}$

4 2000 $\overset{°}{A}$

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Solution

1 3000 A°

2 6000 A°

3 4000 A°

4 2000 A°

Ans. (1) For A to C : EA–EC=hc2000 A° ... (i) and for B to C : EB–EC=hc6000 A° ... (ii) Now for A to B : EA–EB=(EA–EC)–(EB–EC) hcλAB=hc2000–hc6000 1λAB=13000 A° ⇒ λAB=3000 A°

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1 3000 $\overset{°}{A}$

2 6000 $\overset{°}{A}$

3 4000 $\overset{°}{A}$

4 2000 $\overset{°}{A}$