Question

$\bold{\huge{\underline{JEE\:MAINS}}}$


Some energy levels of a molecule are shown in the picture, the proportion of wavelength will be the value of $\bf \: {r = \frac{ \lambda_{1}}{ \lambda _{2} }}$


(a) r = 4/3. (b) r = 2/3

(c) r = 3/4. (d) r = 1/3

Answer 1

We know,

hc/lamda = delta(E)

In this diagram,

The electron looses energy and jumps from higher energy level to lower,
Therefore the electron has a better all round stability and we observe wavelength of different lamda on different jumps of electron.

Where,

delta(E) = change in energy

Now ATQ,

For the first wavelength,

Lamda(1) = hc/delta(E)

= hc/{-E-(-2E)}

= hc/E....(1)

For second wavelength,

Lamda(2) = hc/delta(E)

= hc/{-E-(-4E/3)}

= 3hc/E.... (2)

From (1) and (2),

Taking ratios of both the wavelength,

Lamda(1)/lamda(2) = 1/3

So, the correct option is (d).

Answer 2

As we learnt in

Energy emitted due to transition of electron -

\Delta E= Rhcz^{2}\left ( \frac{1}{n_{f}\, ^{2}}-\frac{1}{n_{i}\, ^{2}} \right )

\frac{1}{\lambda }= Rz^{2}\left ( \frac{-1}{n_{i}\, ^{2}}+\frac{1}{n_{f}\, ^{2}} \right )

- wherein

R= R hydberg\: constant

n_{i}= initial state \\n_{f}= final \: state

From the figure.

\lambda_{1}=\frac{hc}{\Delta E_{1}}=\frac{hc}{-E-(-2t)}=\frac{hc}{E}

\lambda_{2}=\frac{hc}{-E-\left(-\frac{4E}{3} \right )}-\frac{hc}{(\frac{E}{3})}.3\frac{hc}{E}

\frac{\lambda_1}{\lambda_2}=\frac{1}{3}

Correct option is 4

Option 1)

r= \frac{4}{3}

This is an incorrect option.

Option 2)

r= \frac{2}{3}

This is an incorrect option.

Option 3)

r= \frac{3}{4}

This is an incorrect option.

Option 4)

This is the correct option.