Solveeit Logo
Login

Question

Question: The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. Th...

The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be

A

3.4 Ev

B

13.6 eV

C

54.4 Ev

D

122.4 eV

Answer

54.4 Ev

Explanation

Solution

Using 1λ=RZ2(1n12−1n22)\frac{1}{\lambda} = RZ^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)

⇒ 1108.5×10−9=1.1×107×Z2(122−152)\frac{1}{108.5 \times 10^{- 9}} = 1.1 \times 10^{7} \times Z^{2}\left( \frac{1}{2^{2}} - \frac{1}{5^{2}} \right)

⇒ 1108.5×10−9=1.1×107×Z2×21100\frac{1}{108.5 \times 10^{- 9}} = 1.1 \times 10^{7} \times Z^{2} \times \frac{21}{100}

⇒ Z2=100108.5×10−9×1.1×10−7×21=4Z^{2} = \frac{100}{108.5 \times 10^{- 9} \times 1.1 \times 10^{- 7} \times 21} = 4 ⇒ Z = 2

Now Energy in ground state

E=−13.6Z2eV=−13.6×22eV=−54.4eVE = - 13.6Z^{2}eV = - 13.6 \times 2^{2}eV = - 54.4eV