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

The wavelength of the radiation emitted when in a H atom, the electron falls from infinity to stationary state (n=1), is:

1. $15$ $nm$

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 

The energy associated with the fifth orbit of a hydrogen atom is :

$1.$ $-2.18$ $\times$ ${10}^{-18}$ $\mathrm{J}$
$2.$ $-8.72$ $\times$ ${10}^{-20}$ $\mathrm{J}$
$3.$ $-3.88$ $\times$ ${10}^{-21}$ $\mathrm{J}$
$4.$ $-8.72$ $\times$ ${10}^{-19}$ $\mathrm{J}$

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. ${\mathrm{n}}_1$ $=$ $3$ $\mathrm{to}$ ${\mathrm{n}}_2$ $=$ $4$

2. ${\mathrm{n}}_2$  = 3 to n₁ = 2

3. ${\mathrm{n}}_2$  = 3 to n₁ = 1

4. ${\mathrm{n}}_2$  = 2 to n₁ = 1

The electronic transition in the hydrogen atom that emits maximum energy is:

1. 2 $\to$ 1

2. 1 $\to$ 4

3. 4 $\to$ 3

4. 3 $\to$ 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)$\left(\frac{1}{3^2}-\right)$ $\frac{1}{{\mathrm{n}}^2}$. The value of n if the transition is observed at 1285 nm is :