The radii of two of the first four Bohr’s orb | vedclass

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

Question

(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$. Given the ratio of radii is $\frac{r_{n_1}}{r_{n_2}} = \frac{n_1^2}{n_2^2} = \f

Vedclass · Accepted Answer

Either $2.55 \ eV$ or $10.2 \ eV$

Vedclass · Answer

(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$. Given the ratio of radii is $\frac{r_{n_1}}{r_{n_2}} = \frac{n_1^2}{n_2^2} = \frac{1}{4}$,which implies $\frac{n_1}{n_2} = \frac{1}{2}$. Since the orbits are among the first four $(n = 1, 2, 3, 4)$,the possible pairs $(n_1, n_2)$ are $(1, 2)$ or $(2, 4)$. The energy of the $n^{th}$ orbit is $E_n = -\frac{13.6}{n^2} \ eV$. For the pair $(1, 2)$,the energy difference is $\Delta E = E_2 - E_1 = -\frac{13.6}{4} - (-13.6) = -3.4 + 13.6 = 10.2 \ eV$. For the pair $(2, 4)$,the energy difference is $\Delta E = E_4 - E_2 = -\frac{13.6}{16} - (-\frac{13.6}{4}) = -0.85 + 3.4 = 2.55 \ eV$.

The radii of two of the first four Bohr’s orbits of the hydrogen atom are in the ratio $1 : 4$. The energy difference between them may be

Similar Questions

In a hydrogen atom,an electron is transferred from an orbit of radius $1.3225 \ nm$ to another orbit of radius $0.2116 \ nm$. What is the energy (in $J$) of the emitted radiation?

$A$ $100 \, W$ bulb emits monochromatic light of wavelength $400 \, nm$. Calculate the number of photons emitted per second by the bulb.

What is the angular momentum of an electron when it revolves in orbits?

How many spectral lines are obtained when an electron transitions from the $8^{th}$ orbit to the $1^{st}$ orbit in a hydrogen atom?

$1$ mole of photons,each of frequency $2500 \ s^{-1}$,would have approximately a total energy of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module