Question 31:
easyAssertion (A): In solid each electron will have a different energy level.
Reason (R): In solid crystal each electron has a unique position and no two electrons see exactly the same pattern of surrounding charges.
Due to the Pauli exclusion principle, no two electrons can occupy the same quantum state. In a solid, each electron experiences a unique electrostatic environment. Thus, Assertion (A) is true, and Reason (R) provides the correct explanation for it.
Question 32:
easyThe wavelength of Lyman series of hydrogen atom appears in
Lyman series corresponds to transitions to the ground state (\(n=1\)). These high-energy transitions emit radiation in the ultraviolet region of the spectrum.
Question 33:
easyAn electron jumps from orbit \(n = 4\) to \(n = 3\) in hydrogen atom. Wavelength of the emitted radiation is (\(R\) is Rydberg’s constant)
Using Rydberg's formula, \(\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\). Substituting \(n_1 = 3\) and \(n_2 = 4\) gives \(\frac{1}{\lambda} = R \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}\). Hence, \(\lambda = \frac{144}{7R}\).
Question 34:
easyThe ionisation potential of hydrogen is 13.6 V. The energy required to remove an electron from the third orbit of hydrogen is
The energy of an electron in the \(n\)-th orbit of hydrogen is given by \(E_n = -frac{13.6}{n^2}\text{ eV}\). For \(n=3\), \(E_3 = -frac{13.6}{9} = -1.51\text{ eV}\). Thus, the energy required to remove it is 1.51 eV.
Question 35:
easyIn a hypothetical situation, all the atoms in a hydrogen sample are excited to same state. During de-excitation, photon with lowest energy was found to have \(0.66\text{ eV}\). The photon with the highest energy will have energy equal to
For hydrogen atom, \(E_n - E_{n-1} = 0.66\text{ eV}\) corresponds to \(n = 5\) to \(n = 4\) transition (since \(E_5 - E_4 = -0.85 - (-1.51) = 0.66\text{ eV}\)). The highest energy photon is emitted for transition from \(n = 5\) to \(n = 1\), which is \(E_5 - E_1 = -0.85 - (-13.6) = 12.75\text{ eV}\).
Question 36:
easyThe wavelength of Balmer series of hydrogen atom appears in
The transitions in the Balmer series end on \( n = 2 \). The wavelengths of these transitions lie in the range of 380 nm to 700 nm, which belongs to the visible region of the electromagnetic spectrum.
Question 37:
easyAn electron in a hydrogen atom makes a transition from \(n = n_1\) to \(n = n_2\). The time period of revolution of the electron in the initial state is eight times that in final state. The possible value of \(n_1\) and \(n_2\) are
The orbital period is proportional to \(n^3\). Since \(T_1 = 8 T_2\), we must have \(n_1^3 = 8 n_2^3\), which gives \(n_1 = 2n_2\). Thus, \(n_1 = 4\) and \(n_2 = 2\) is correct.
Question 38:
easyWhen an $\alpha$ particle of mass m moving with velocity v bombards on a heavy nucleus of charge ‘Ze’, its distance of closest approach from the nucleus depends on mass:
(2016 – I)
At the distance of closest approach $r_0$, kinetic energy is converted to potential energy. $\frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$. Rearranging gives $r_0 = \frac{4 Z e^2}{4\pi\epsilon_0 m v^2}$, which shows $r_0 \propto \frac{1}{m}$.
Question 39:
easyAn alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to:
(2010 Pre)
The distance of closest approach is found by equating kinetic energy to electrostatic potential energy: $K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$. Since $K = \frac{1}{2} m v^2$, we have $r_0 \propto \frac{1}{m}$.
Question 40:
easyIn a Rutherford scattering experiment, when a projectile of charge $z_1$ and mass $M_1$ approaches a target nucleus of charge $z_2$ and mass $M_2$, the distance of closest approach is $r_0$. The energy of the projectile is
(2009)
At the distance of closest approach $r_0$, the entire kinetic energy of the projectile is converted into electrostatic potential energy. Energy $E = \frac{1}{4\pi\epsilon_0} \frac{z_1 z_2}{r_0}$. Thus, the energy is directly proportional to the product of charges $z_1 z_2$.