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 |