The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a 0 , where a 0 is the Bohr radius. Its orbital angular momentum is 3 h 2 π . It is given that h is Planck's constant and R is Rydberg constant. the

Correct answer: A. 9 3 2 R R _ n = 4.5 a _ 0 ⇒ a _ 0 n ^ 2 / z = 4.5 a _ 0 ∴ n = 3 , z = 2 L = mvr = 3 h 2 π [as n = 3 , z = 2] (1)/(λ) = Rz ^ 2 ( 1 n _ f ^ 2 - 1 n _ 1 ^ 2 ) 1 λ _ 3 → 1 = R 4 [(1)/(1) - (1)/(9)] = 4 R (8)/(9) ⇒ λ _ 3 → 1 = 9 3 2 R 1 λ _ 2 → 1 = R 4 [(1)/(1) - (1)/(4)] = (3)/(4) 4 R ⇒ λ _ 2 → 1 = 1 3 R 1 λ _ 3 → 2 = R 4 [(1)/(4) - (1)/(9)] = (5)/(3 6) 4 R ⇒ λ _ 3 → 2 = 9 5 R

JEE Advanced 2013Paper 2

Exam: JEE Advanced
Year: 2013
Paper: Paper 2
Subject: Physics
Chapter: Atomic Physics

JEE Advanced 2013 · Paper 2 · Physics · Atomic Physics

The radius of the orbit of an electron in a Hydrogen-like atom is...

The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a₀, where a₀ is the Bohr radius. Its orbital angular momentum is

\dfrac{3 \text{h}}{2 \text{\pi}}.

It is given that h is Planck's constant and R is Rydberg constant. the possible wavelength (s), when the atom de-excites, is (are) The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a 0 , where a 0 is the Bohr radius. Its orbital

The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a 0 , where a 0 is the Bohr radius. Its orbital

Options

A $\dfrac{9}{3 2 \text{R}}$
B $\dfrac{9}{1 6 \text{R}}$
C $\dfrac{9}{5 \text{R}}$
D $\dfrac{4}{3 \text{R}}$

Correct answer

A. $\dfrac{9}{3 2 \text{R}}$

Step-by-step solution

{\text{R}}_{\text{n}} = \text{4.5} {\text{a}}_{0} ⇒ {\text{a}}_{0} {\text{n}}^{2} / \text{z} = \text{4.5} {\text{a}}_{0}

∴ \text{n} = 3 , \text{z} = 2

\text{L} = \text{mvr} = \dfrac{3 \text{h}}{2 \pi} \left[\text{as n} = 3 \text{,} \text{z} = 2\right]

\dfrac{1}{\lambda} = {\text{Rz}}^{2} \left(\dfrac{1}{{\text{n}}_{\text{f}}^{2}} - \dfrac{1}{{\text{n}}_{\text{1}}^{2}}\right)

\dfrac{1}{{\lambda}_{3 \to 1}} = \text{R} 4 \left[\dfrac{1}{1} - \dfrac{1}{9}\right] = 4 \text{R} \dfrac{8}{9} ⇒ {\lambda}_{3 \to 1} = \dfrac{9}{3 2 \text{R}}

\dfrac{1}{{\lambda}_{2 \to 1}} = \text{R} 4 \left[\dfrac{1}{1} - \dfrac{1}{4}\right] = \dfrac{3}{4} 4 \text{R} ⇒ {\lambda}_{2 \to 1} = \dfrac{1}{3 \text{R}}

\dfrac{1}{{\lambda}_{3 \to 2}} = \text{R} 4 \left[\dfrac{1}{4} - \dfrac{1}{9}\right] = \dfrac{5}{3 6} 4 \text{R} ⇒ {\lambda}_{3 \to 2} = \dfrac{9}{5 \text{R}}

Practice Atomic Physics on Quantrex Academy →

More from Atomic Physics

Consider a hydrogen atom with v_k , r_k , and K_k denoting the velocity, orbital radius… JEE Advanced 2026 The ratio of momentum of the photons of the 1^ st and 2^ nd line of Balmer series of… JEE Main 2026 In the hydrogen atom, the electron makes a transition from the higher orbit ( i ) to a… JEE Main 2026 In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound… JEE Main 2026 Using Bohr's model, calculate the ratio of the magnetic fields generated due to the… JEE Main 2026 Angular momentum of an electron in a hydrogen atom is (3h)/(π) , then the energy of the… JEE Main 2026 Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10^ 5 ~m /… JEE Main 2026 In hydrogen atom spectrum, (R arrow Rydberg's constant ) A. the maximum wavelength of the… JEE Main 2026 Full Atomic Physics list All JEE Advanced PYQs