Question 1:
easyThe ground state energy of hydrogen atom is \(-13.6\text{ eV}\). The energy needed to ionize hydrogen atom from its second excited state will be
Second excited state corresponds to \(n = 3\). The energy is \(E_3 = -\frac{13.6}{3^2} = -1.51\text{ eV}\). The ionization energy required is \(E_{\text{ion}} = 0 - E_3 = 1.51\text{ eV}\).
Question 2:
easyThe wavelength of Lyman series of hydrogen atom appears in
The Lyman series transitions terminate at the ground state (\(n = 1\)). The photon energies emitted in these transitions correspond to the ultraviolet region of the electromagnetic spectrum.
Question 3:
easyAn electron in the hydrogen atom jumps from state \(n\) to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function \(4.09\text{ eV}\). If the stopping potential of the photoelectrons is \(8\text{ V}\), the value of \(n\) is
Formula: \(E = K_{max} + \Phi_0 = 8 + 4.09 = 12.09\text{ eV}\). For hydrogen atom transition, \(13.6 \left(1 - \frac{1}{n^2}\right) = 12.09\). Solving this gives \(n = 3\).
Question 4:
easyConsider a hypothetical hydrogen like atom. Wavelength of spectral lines corresponding to transition of electron from \( n = m \) to \( n = 1 \) are given by \( \lambda = \frac{1800 m^2}{m^2 – 1} \text{ \A^\circ} \) \( (m = 2, 3, 4, 5, \dots, \infty) \). Minimum energy of emitted photon is nearly
Minimum energy corresponds to the maximum wavelength, which occurs at \( m = 2 \). Thus, \( \lambda_{\max} = \frac{1800 \times 2^2}{2^2 - 1} = 2400 \text{ \A^\circ} \). The minimum energy is \( E_{\min} = \frac{12400}{\lambda_{\max}} = \frac{12400}{2400} \approx 5.17 \text{ eV} \approx 5.2 \text{ eV} \).
Question 5:
easyThe wave number of a photon in bracket series of hydrogen atom is \(\frac{9}{400}R\). The electron has undergone transition from the orbit having quantum number
For Brackett series, \(n_1 = 4\). The wave number formula is \(bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\). Thus, \(frac{9}{400}R = R \left( \frac{1}{16} - \frac{1}{n_2^2} \right) ⇒ \frac{1}{n_2^2} = \frac{1}{16} - \frac{9}{400} = \frac{16}{400} = \frac{1}{25} ⇒ n_2 = 5\).
Question 6:
easyFor H-atom, Lyman spectral lines must lie in
The Lyman series corresponds to transitions to the ground state (\(n = 1\)) in a hydrogen atom, and the emitted photons lie in the ultraviolet region.
Question 7:
easyIf the energy of a hydrogen atom in \(n^{text{th}}\) orbit is \(E\), then energy in the \(n^{text{th}}\) orbit of a singly ionized helium atom will be
The energy in the \(n^{text{th}}\) orbit is proportional to \(Z^2\), where \(Z\) is the atomic number. For hydrogen, \(Z=1\) and for helium, \(Z=2\). Therefore, the energy of singly ionized helium is \(2^2 = 4\) times the energy of hydrogen, which is \(4E\).
Question 8:
easyFor H-atom, Lyman spectral lines must lie in:
The Lyman series corresponds to transitions from outer orbits to the first orbit (\(n=1\)) of hydrogen. These emissions are of high energy and fall in the ultraviolet region.
Question 9:
easyAssertion (A): If the accelerating potential of a X-Ray tube is increased then the characteristic wavelength decreases.
Reason (R): The cut-off wavelength for a X-Ray tube is given by \(\lambda_{\text{min}} = \frac{hc}{eV}\), where \(V\) is accelerating potential.
Characteristic wavelength depends on target material, not accelerating potential. However, if 'characteristic wavelength' in (A) refers to 'cut-off wavelength', then \(\lambda_{\text{min}} = hc/(eV)\) implies increasing \(V\) decreases \(\lambda_{\text{min}}\). Under this interpretation, (A) is true, and (R) is true and explains (A).
Question 10:
easyAssertion (A): Cut-off wavelength of x-ray is independent of type of target metal
Reason (R): Wavelength of \(K_{alpha}\) x-ray depends upon type of target metal.
Concept: X-ray production mechanisms.
Formula: Cut-off wavelength \(\lambda_{\text{min}} = \frac{hc}{eV}\) (depends on voltage). Characteristic X-ray energy \(E = h\nu\) (depends on atomic transitions).
Solution: The cut-off (minimum) wavelength of continuous X-rays depends only on the accelerating voltage, not the target material. So (A) is true. \(K_{\alpha}\) X-rays are characteristic X-rays, whose wavelengths are specific to the target material. So (R) is true. However, (R) explains characteristic X-rays, not the cut-off wavelength; thus, (R) is not a correct explanation for (A).