Question 121:
easyHydrogen atoms are excited from ground state of the principle quantum number 4. Then the number of spectral lines observed will be:
(1993)
The number of possible spectral lines emitted when transitioning from the nth state to the ground state is $\frac{n(n-1)}{2}$. For $n=4$, the number of lines is $\frac{4 \times 3}{2} = 6$.
Question 122:
easyWhich source is associated with a line emission spectrum?
(1993)
Line emission spectra are characteristic of excited atoms in low-pressure gases. A neon street sign contains low-pressure neon gas which, when excited, emits a characteristic line spectrum.
Question 123:
easyIn terms of Bohr radius $a_0$, the radius of the second Bohr orbit of a hydrogen atom is given by:
(1992)
The radius of the nth Bohr orbit for a hydrogen atom is $r_n = a_0 n^2$. For the second orbit ($n=2$), the radius is $r_2 = a_0 (2^2) = 4a_0$.
Question 124:
easyEnergy E of a hydrogen atom with principal quantum number n is given by $E = \frac{-13.6}{n^2} eV$. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately:
(2004)
The energy of the emitted photon is $\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) eV$. This gives $\Delta E = 13.6 \times \frac{5}{36} = 1.88 eV \approx 1.9 eV$.
Question 125:
easyThe ionisation energy of hydrogen atom is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between 3rd and 4th orbit is
(1992)
The energy of the nth orbit is $E_n = -\frac{13.6}{n^2} eV$. For $n=3$, $E_3 = -1.51 eV$, and for $n=4$, $E_4 = -0.85 eV$. The energy difference is $\Delta E = E_4 - E_3 = -0.85 - (-1.51) = 0.66 eV$.
Question 126:
easyThe Bohr model of atoms:
(2004)
According to Bohr's second postulate, the electron revolves only in those orbits for which its angular momentum is an integral multiple of $h/(2\pi)$. Hence, it assumes that the angular momentum of electrons is quantized.
Question 127:
easyThe ground state energy of H-atom is 13.6 eV. The energy needed to ionize H-atom from its second excited state is:
(1991)
The second excited state corresponds to $n=3$. The energy of this state is $E_3 = -\frac{13.6}{3^2} = -1.51 eV$. The energy required to remove the electron to infinity (ionization) is $0 - (-1.51) = 1.51 eV$.
Question 128:
easyIn which of the following systems will be radius of the first orbit (n = 1) be minimum:
(2003)
The radius of the nth orbit in a hydrogen-like species is given by $r_n \propto \frac{n^2}{Z}$. For the first orbit ($n=1$), the radius is inversely proportional to the atomic number $Z$. Doubly ionised lithium ($Li^{2+}$) has the maximum $Z=3$, thus it has the minimum radius.
Question 129:
easyThe energy of hydrogen atom in $n^{th}$ orbit is $E_n$ then the energy in $n^{th}$ orbit of singly ionised helium atom will be:
(2001)
Energy of an electron in a hydrogen-like atom is $E \propto Z^2$. For hydrogen $Z=1$, $E = E_n$. For singly ionised helium ($He^+$), $Z=2$. Therefore, the energy in the same orbit is $2^2 E_n = 4 E_n$.
Question 130:
easyMaximum frequency of emission is obtained for the transition:
(2000)
Frequency $\nu \propto \Delta E$. Emission occurs when transitioning from a higher to a lower energy state. The energy difference between $n=2$ and $n=1$ ($10.2 eV$) is the largest among the given emission transitions.