AIIMS Physics Structure Of Atom Class 11 Questions

If energy of electron in ground state is^{-13}.6 then find out speed of electron in fourth orbit of H⁻atom

15 eV is given to e⁻ in 4^{th} orbit then find it's final energy when it comes out of H⁻atom

An electron is revolving in n = 3 orbit. What will be the magnetic field at the centre of hydrogen atom?

An electron collides elastically with H⁻like atom and excites it from ground state to n = 3. Find out the energy transfer to H⁻like atom

The radius of electron in excited state is given by, r = r₀ (n²/Z) For the radius to be equal to radius of hydrogen atoms, r = r₀. Therefore, r₀ = r₀ (n²/Z) n = √Z = √4 = 2 Thus, at first excited state of Be³⁺ radius of e⁻ will be same as H atoms and electron in ground state.

Assertion: For revolving electron, direction of angular momentum and magnetic moment are opposite. Reason: Charge of electron is negative.

The expression of energy is given by, E = -13.6 * (z²/n²). Substitute the values. E = -13.6 * (2²/2²) = -13.6 eV

Assertion: For an element generally N ≥ Z (N=number of neutrons, Z= atomic number) Reason: Neutrons always experience attractive nuclear force.

Assertion: Hydrogen atom consists of only one electron but its emission spectrum has many lines. Reason: Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.

The electron rises to some higher energy level when the atom gets some energy from external (outside). The electron can return either, directly to the lower energy level after passing through other lower energy levels, hence all possible transitions take place in the source and many lines are seen in the spectrum.

Find out the velocity of electron in second orbit of helium.

Calculate λ from the given expression. 1/λ = R·z² (1/(2)² - 1/(5)²) 1/λ = 1.09×10⁷ (1)² (1/4 - 1/25) λ = 4.37×10⁻⁷ Calculate the ground state energy of hydrogen atom. E_n = hc/λ = (4.135×10⁻¹⁵ eV)(3×10⁸ m/s) / 4.37×10⁻⁷ = 2.83 eV Binding Energy is,

The shortest wavelength present in the paschen series of spectral lines are, 1/λₘᵢₙ = R(1)²[1/(3)² - 1/(∞)²] 1/λₘᵢₙ = R/9 λₘᵢₙ = 9/1.0965×10⁻³ λₘᵢₙ = 8208 Å

The Bohr radii of a hydrogen like atom are r₁ and r₂. Find the wavelength of photon when electron jumps from r₂ to r₁.

For first line of Balmer series, the energy is, \newline E_1 = 13.6 (2)^2 \left(\frac{1}{4} - \frac{1}{9}\right) \newline The energy required to eject electron from n = 2 level in H atom is calculated as, \newline E = 13.6 \left[\left(¹⁻ \frac{4}{9}\right) - \left(\frac{1}{4}\right)\right] \newline = 13.6 \left(\frac{11}{36}\right) \newline = 4.155 eV

The wavelength of first spectral line in the Balmer series of hydrogen atom is 6534 Å. \frac{1}{λ} = RZ² \left[ \frac{1}{n₁²} - \frac{1}{n₂²} \right] \frac{1}{6561} = R(1)² \left[ \frac{1}{2²} - \frac{1}{3²} \right] ...(1) For the second spectral line in the Balmer series of singly ionised helium ion, \frac{1}{λ} = R(2)² \left[ \frac{1}{2²} - \frac{1}{4²} \right] ...(2)

According to postulates of Bohr’s atomic model, the electron revolves around the nucleus in fixed orbit of definite radii. As long as the electron is in a certain orbit it does not radiate any energy.

Rutherford confirmed the repulsive force on α-particle due to nucleus varies with distance according to inverse square law and that the positive changes are concentrated at the centre and not distributed throughout the atom.

In the following questions, a statement of assertion is given followed by a corresponding statement of reason. Assertion:(A) Hydrogen atom consists of only one electron but its emission spectrum has many lines. Reason:Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.

Assertion: The positive charged nucleus of an atom has a radius of 10⁻¹⁵ m. Reason: In α-particle scattering experiment, the distance of closest approach for α particle is 10⁻¹⁵ m.

Assertion: The positively charged nucleus of an atom has a radius of almost 10⁻¹⁵ m. Reason: In α particle scattering experimenting, the distance of closest approach for α-particles is ≈ 10⁻¹⁵ m.

Assertion : Electrons in the atom are held due to coulomb forces. Reason : The atom is stable only because the centripetal force due to Coulomb’s law is balanced by the centrifugal force.

Assertion (A) The magnetic moment (μₗ) of an electron revolving around the nucleus decreases with increasing principle quantum number (n). Reason (R) Magnetic moment of the revolving electron, μ ∝ n.