Home
Class 12
PHYSICS
An electron jumps from the 4^(th) orbit ...

An electron jumps from the `4^(th)` orbit to the `2^(nd)` orbit of hydrogen atom. Given the Rydberg's constant `R=10^5cm^-1`. The frequency in Hz of the emitted radiation is

A

`3/16xx10^5` Hz

B

`3/16xx10^15` Hz

C

`9/16xx10^15` Hz

D

`3/4xx10^15` Hz

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

When an electron falls from 4th orbit to 2nd orbit in hydrogen atom, the spectral line is observed in

When an electron falls from higher orbit to third in hydrogen atom, the spectral line observed

The energy of an electron in the 1st Bohr orbit of hydrogen atom is

Find the energy of the electron in eV in the third Bohr orbit of the hydrogen atom. ( Rydberg's constant R = 1.097 xx10^7 m^-1 Planck's constant (h) = 6.63xx10^-34 J.S , velocity of light in air (c) = 3xx 10^8 m//s)

The wave number of first line in Balmer series of hydrogen spectrum is (Rydberg constant, R_H = 109,678 cm^-1 ) nearly

The radius of the first orbit of hydrogen atom is 0.52 xx 10^-8 cm . The radius of the first orbit of He^+ ion is

Find the value of Rydberg's constant if the energy of electron in second orbit in hydrogen atom is -3.4eV

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10–16 s. The frequency of revolution of the electron in its first excited state (in s–1) is:

If the radius of 1st Bohr orbit in hydrogen atom is 0.5 A^@ 'then radius of 3rd Bohr orbit is

Starting from the formula for energy of an electron in the nth orbit of hydrogen atom, derive the formula for the wavelength of spectral lines.