JEE Physics Atoms MCQs Set 03

Answer: (a) \( -43.52\times10^{-19} \) J

Explanation:
1. For an electron in a Bohr orbit, \( U = 2E \) (since \( K = -E \) and \( U = -2K \))
2. \( U = 2\times(-21.76\times10^{-19}) = -43.52\times10^{-19} \) J. Answer: (a)

Teacher's Note:
1. Potential energy is twice the total energy and kinetic energy equals minus the total energy.
2. All three energies are in the ratio K : U : E = 1 : -2 : -1.

Answer: (d) 656 nm

Explanation:
1. Longest wavelength corresponds to the transition \( n = 3 \to 2 \)
2. \( \frac1\lambda = R\left(\frac14 - \frac19\right) = \frac{5R}{36} \)
3. \( \lambda = \frac{36}{5R} = \frac{36}{5\times1.097\times10^7} \approx 656 \) nm. Answer: (d)

Teacher's Note:
1. The longest wavelength is the lowest energy transition in the series.
2. Balmer lines end on n = 2.

Answer: (c) Time period of rotation is increased while angular momentum is decreased.

Explanation:
1. Going from n = 1 to n = 2, the radius increases, kinetic energy decreases, potential energy increases and total energy increases
2. Angular momentum \( L = \frac{nh}{2\pi} \) increases, not decreases
3. The time period increases and the angular speed decreases. So the statement in (c) is incorrect. Answer: (c)

Teacher's Note:
1. Angular momentum in the Bohr model is proportional to n.
2. Higher orbits have longer time periods and smaller angular speeds.

Answer: (b) \( \frac{ke^2}{2}\left[\frac{1}{R_2}-\frac{1}{R_1}\right] \)

Explanation:
1. Total energy in a circular orbit \( E = -\frac{ke^2}{2r} \)
2. Decrease \( = E_1 - E_2 = -\frac{ke^2}{2R_1} + \frac{ke^2}{2R_2} = \frac{ke^2}{2}\left[\frac{1}{R_2} - \frac{1}{R_1}\right] \)
3. This is positive when \( R_2 < R_1 \), as the electron drops to an inner orbit. Answer: (b)

Teacher's Note:
1. Total energy equals minus half the potential-energy magnitude for a circular orbit.
2. The decrease is the initial energy minus the final energy.

Answer: (c) 10.2 eV

Explanation:
1. Ground-state atoms can absorb only energies equal to level differences \( 13.6\left(1-\frac{1}{n^2}\right) \) eV
2. The smallest is \( n = 1 \to 2 \): \( 13.6\times\frac34 = 10.2 \) eV
3. Only 10.2 eV matches the options. Answer: (c)

Teacher's Note:
1. Absorption by an atom needs the photon energy to equal an energy-level gap.
2. The first gap from n = 1 is 10.2 eV.