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 ionization energy for H atom in the ground state is \(2.18 \times10^{-18}~ \mathrm J.\) The process energy requirements will be: \(( He^+(g) \rightarrow He^{2+}(g) + e^- )\)

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 :

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

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

\(1 .   1 . 52   \times   10^{6}   m^{- 1}\)
\(2 .   3 . 14   \times   10^{6}   \left(cm\right)^{- 1}\)
\(3 .   15 . 2   \times   10^{6}   m^{- 1}\)
\(4 .   1 . 52   \times   10^{6}   \left(cm\right)^{- 1}\)

The energy of an electron in an H - atom is given by \(E_n=(-2.18 \times10^{-18})/n^2~\mathrm J.\) The shortest wavelength of light that can be used to remove an electron completely from \(n = 2\) orbit will be:

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 :

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