The wave numbers of three spectral lines of H | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the Balmer series, the wave number $\bar{v}$ is given by the formula: $\bar{v} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$, where $n = 3

Vedclass · Accepted Answer

$ \frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100} $

Vedclass · Answer

(A) For the Balmer series, the wave number $\bar{v}$ is given by the formula: $\bar{v} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$, where $n = 3, 4, 5, \dots$ For $n = 3$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36}$ For $n = 4$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left[ \frac{4-1}{16} \right] = \frac{3R}{16}$ For $n = 5$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{25} \right] = R \left[ \frac{25-4}{100} \right] = \frac{21R}{100}$ Thus, the set of spectral lines is $\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}$.

The wave numbers of three spectral lines of $H$ atom are considered. Identify the set of spectral lines belonging to the Balmer series. ($R =$ Rydberg constant)

Similar Questions

The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength in Balmer series of $He^{+}$ is

For a $H$ atom,the wavelength for the highest energy transition is $91.2 \ nm$. Calculate the corresponding wavelength for $He^{+}$ in $nm$.

$A$ bulb emits light of wavelength $4500 \ \mathring{A}$. The bulb is rated at $150 \ W$ and $8 \ \%$ of the energy is emitted as light. What is the number of photons emitted by the bulb in $1 \ s$?

If $P.E.$ of $3^{rd}$ shell is $-20 \ eV/atom$,then find the $I.P.$ of the $3^{rd}$ shell of that element.

Which electron transition in a hydrogen atom requires the largest amount of energy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module