The 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 \(E_{\text{ion}} = 0 - E_3 = 1.51\text{ eV}\).
The 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.
An 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 \(n = 3\).
Consider 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 \( E_{\min} = \frac{12400}{\lambda_{\max}} = \frac{12400}{2400} \approx 5.17 \text{ eV} \approx 5.2 \text{ eV} \).
The 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, \(frac{9}{400}R = R \left( \frac{1}{16} - \frac{1}{n_2^2} \right) ⇒ \frac{1}{n_2^2} = \frac{1}{16} - \frac{9}{400} = \frac{16}{400} = \frac{1}{25} ⇒ n_2 = 5\).
For 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.
If 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 \(4E\).
For 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.
Assertion (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).
Assertion (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).