Determine the maximum number of emission lines produced when an electron in a hydrogen atom transitions from the n = 6 energy level to the ground state.
1. 30
2. 21
3. 15
4. 28
The ratio of the wavelengths of the last lines of the Balmer to Lyman series is
1. | 4:1 | 2. | 27:5 |
3. | 3:1 | 4. | 9:4 |
The wavelength of the radiation emitted when in a H atom, the electron falls from infinity to stationary state (n=1), is:
1.
2. 192 nm
3. 406 nm
4. 91 nm
When an electron jumps from n=5 to n=1 in a hydrogen atom, the number of spectral lines obtained is
1. | 3 | 2. | 4 |
3. | 6 | 4. | 10 |
The energy associated with the fifth orbit of a hydrogen atom is :
The wavelength of light emitted when the electron in a H atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2, is :
1. 586 mm
2. 486 nm
3. 523 nm
4. 416 pm
The transition in the hydrogen spectrum that would have the same wavelength as Balmer transition from n = 4 to n = 2 of He+ spectrum is -
1.
2. = 3 to n1 = 2
3. = 3 to n1 = 1
4. = 2 to n1 = 1
The electronic transition in the hydrogen atom that emits maximum energy is:
1. 2 1
2. 1 4
3. 4 3
4. 3 2
The maximum wavelength in the Lyman series of He+ ion is-
1. 3R
2. 1/3R
3. 1/R
4. 2R
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) . The value of n if the transition is observed at 1285 nm is :
1. | 6 | 2. | 5 |
3. | 8 | 4. | 9 |