HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q12 Solution: A hydrogen atom in a state having a binding energy of 0.85 eV…

QPages 384-386: Q12: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q12 Solution: A hydrogen atom in a state having a binding energy of 0.85 eV…

A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 eV. a) Identify the quantum numbers

n

of the upper and the lower energy states involved in the transition. b) Find the wavelength of the emitted radiation.

Step-by-Step Verified Solution

**Answer:** (a) Upper state n = 4, lower state n = 2. (b) Wavelength = 486 nm. **Step-by-Step Explanation:** (a) Binding energy 0.85 eV:

E_n = -0.85 \Rightarrow 13.6/n^2 = 0.85 \Rightarrow n^2 = 16 \Rightarrow n = 4

. Excitation energy 10.2 eV:

E_n = -13.6 + 10.2 = -3.4 \Rightarrow 13.6/n^2 = 3.4 \Rightarrow n^2 = 4 \Rightarrow n = 2

. So the transition is 4 _ 2. (b)

\frac{1}{\lambda} = R(1/2^2 - 1/4^2) = R(1/4 - 1/16) = 3R/16

\lambda = 16/(3 \times 1.097 \times 10^7) = 4.86 \times 10^{-7}

m = 486 nm [citation:3].

QPages 384-386: Q12: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q12 Solution: A hydrogen atom in a state having a binding energy of 0.85 eV…

Step-by-Step Verified Solution

All Questions in this Chapter

Related Learning Resources & Exam Preparation