The electron in a hydrogen atom is in the fir | vedclass

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

Question

(A) According to Bohr's energy formula,$\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ 	ext{eV/atom}$.
For the transition fr

Vedclass · Accepted Answer

$32 / 27$

Vedclass · Answer

(A) According to Bohr's energy formula,$\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ 	ext{eV/atom}$.
For the transition from $n=1$ to $n=3$:
$\Delta E_{1}$ ${ightarrow 3} = 13.6 \left[ \frac{1}{1^2} - \frac{1}{3^2} ight] = 13.6 \left[ 1 - \frac{1}{9} ight] = 13.6 \left[ \frac{8}{9} ight]$.
For the transition from $n=1$ to $n=2$:
$\Delta E_{1}$ ${ightarrow 2} = 13.6 \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = 13.6 \left[ 1 - \frac{1}{4} ight] = 13.6 \left[ \frac{3}{4} ight]$.
Therefore,the ratio is:
$\frac{\Delta E_{1 	o 3}}{\Delta E_{1 	o 2}} = \frac{13.6 \left[ \frac{8}{9} ight]}{13.6 \left[ \frac{3}{4} ight]} = \frac{8}{9} 	imes \frac{4}{3} = \frac{32}{27}$

The electron in a hydrogen atom is in the first Bohr orbit $(n=1)$. The ratio of transition energies,$E(n=1 \rightarrow n=3)$ to $E(n=1 \rightarrow n=2)$,is

Similar Questions

According to Bohr's theory,the angular momentum of an electron in the $5^{th}$ orbit is ....... (in $h/\pi$)

Find the frequency and wavelength of light required for the ionization of the solution. (Ionization energy of the solution $= 8.2 \times 10^{-19} \ J$)

How many photons are required to provide $1 \, J$ of energy with a wavelength of $5000 \, \mathring{A}$?

If the wavelength of the limiting line of the Lyman series of the hydrogen atom is $16$ times the wavelength of the limiting line of the same series for an ion $X$,then what is $X$?

In which transition of the hydrogen atom will the emitted radiation have the lowest frequency?

Vedclass Test Series

Exam Paper Generator

Online Exam Module