The maximum number of emission lines obtained when the excited electron of the H atom jumps from n = 6 to the ground state is -

1. 30

2. 21

3. 15

4. 28

The ionization energy for H atom in the ground state is 2.18 x 10^-18 J. The process energy requirements will be: \(( He^+(g) \rightarrow He^{2+}(g) + e^- )\)
1. 8.72 x 10^-18 J
2. 7.54 x 10^-18 J
3. 5.67 x 10^-17 J
4. 2.18 x 10^-17 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 energy associated with the fifth orbit of a hydrogen atom is 

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

The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen would be -

$$\begin{gathered} 1. 1.52 \times {10}^6 {\mathrm{m}}^{-1} \\ 2. 3.14 \times {10}^6 {\mathrm{cm}}^{-1} \\ 3. 15.2 \times {10}^6 {\mathrm{m}}^{-1} \\ 4. 1.52 \times {10}^6 {\mathrm{cm}}^{-1} \end{gathered}$$

The energy of an electron in H - atom is given by E_n = $\frac{(–2.18 \times {10}^{-18} )}{{\mathrm{n}}^2}$ J.  The shortest wavelength of light that can be used to remove an electron completely from n = 2 orbit will be

$$\begin{gathered} 1. 3647 \stackrel{°}{\mathrm{A}} \\ 2. 5132 \stackrel{°}{\mathrm{A}} \\ 3. 3017 \stackrel{°}{\mathrm{A}} \\ 4. \mathrm{None} \mathrm{of} \mathrm{these} \end{gathered}$$

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)$\left(\frac{1}{3^2}- \frac{1}{{\mathrm{n}}^2}\right)$. The value of n if the transition is observed at 1285 nm is
1. 6
2. 5
3. 8
4. 9

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