Question

Please help me with this question

Options are also provided

Class-11 ​

Answer 1

Answer:B

Explanation:

Hope the file attached clears your doubts

Answer 2

Answer:-

➣

$\bf{Option \: B \implies[486nm]}$

• To Find:-

Wavelength of the Spectral Line

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

• Solution:-

Rydberg's Equation:

${\boxed{\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )}}$

Where,

• $\bf{\lambda}$ = Wavelength of radiation

• $\bf{R_H}$ = Rydberg's Constant

$\\ \implies 1.097\times 10^7m^{-1}$

• $\bf{n_f}$ = Final energy level = 2

• $\bf{n_i}$= Initial energy level = 4

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{4^2} \right )$

$\lambda =\frac{1}{2.056\times 10^6m^{-1}}$

$\implies{\bf{4.862\times 10^{-7}m}}$

Now, Converting into nanometre [nm]:-

$1m=10^9nm$

Therefore,

$\bf{4.862\times 10^{-7}m\times (\frac{10^9nm}{1m})}$

$\large \implies \bf{486.2nm}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬