Question 71:
easyIf \( E \) is the energy of \( n^{\text{th}} \) orbit of hydrogen atom, the energy of \( n^{\text{th}} \) orbit of \( \text{He}^+ \) ion will be
The energy of an electron in a hydrogen-like atom is given by \( E_n \propto Z^2 \). For hydrogen, \( Z = 1 \), and for helium ion \( \text{He}^+ \), \( Z = 2 \). Therefore, \( E_{\text{He}^+} = Z^2 E = 4E \).
Question 72:
easyIn an experiment on photoelectric emission for incident light of wavelength \( 1.98 \times 10^{-7} \text{ m} \), stopping potential is found to be \( 2.5 \text{ V} \). What is maximum kinetic energy of emitted photoelectron?
The maximum kinetic energy of emitted photoelectrons is related to the stopping potential by \( K_{\max} = e V_s \). Given \( V_s = 2.5 \text{ V} \), the maximum kinetic energy is simply \( 2.5 \text{ eV} \).
Question 73:
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 74:
easyThe speed of photons of radiation having wavelength \(\lambda\), in vacuum is proportional to
In vacuum, the speed of all photons (electromagnetic waves) is constant (\(c = 3 \times 10^8 \text{ m/s}\)), which is independent of their wavelength. Thus, speed is proportional to \(\lambda^0\).
Question 75:
easyIf the radius of a nucleus of mass number 2 is R, then radius of a nucleus of mass number 16 is
The radius of a nucleus is given by \(R \propto A^{1/3}\). Therefore, \(frac{R'}{R} = \left(\frac{16}{2}\right)^{1/3} = 8^{1/3} = 2\), so \(R' = 2R\).
Question 76:
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 77:
easyIf the kinetic energy of a particle is increased to 16 times, the percentage decrease in de Broglie wavelength of particle is
de Broglie wavelength is \(lambda = \frac{h}{\sqrt{2mK}}\). If \(K' = 16K\), then \(lambda' = \frac{\lambda}{\sqrt{16}} = \frac{\lambda}{4}\). The percentage decrease is \(\frac{\lambda - \lambda/4}{\lambda} \times 100% = 75%\).
Question 78:
easyThe stopping potential in the photoelectric experiment is 1.6 V. The maximum kinetic energy of photoelectrons emitted is
Maximum kinetic energy of photoelectrons is \(K_{\text{max}} = e V_s = 1.6 \times 1.6 \times 10^{-19}\text{ J} = 2.56 \times 10^{-19}\text{ J}\).
Question 79:
easyChoose the incorrect statement among the following.
Nuclides with the same atomic number \(Z\) but different neutron number \(N\) are called isotopes. Isotones are nuclides with the same number of neutrons.
Question 80:
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 \(4E\).