Question 111:
easyThe transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from
(2012 Mains)
Transition to $n=1$ gives UV radiation. Transition to $n=2$ gives visible light (Balmer series). Transition to $n=3$ (Paschen series) gives infrared radiation. Therefore, the transition $4 \rightarrow 3$ will yield infrared radiation.
Question 112:
easyAn electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be
(2012 Pre)
Momentum of the emitted photon is $p = \frac{h}{\lambda} = hR(\frac{1}{1^2} - \frac{1}{5^2}) = hR(\frac{24}{25})$. By conservation of momentum, the recoil momentum of the atom is $mv = p = \frac{24hR}{25}$. Thus, velocity $v = \frac{24hR}{25m}$.
Question 113:
easyElectron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength $\lambda_1 : \lambda_2$ emitted in the two cases is:
(2012 Pre)
Third excited state is $n=4$, second is $n=3$, first is $n=2$. For $4 \rightarrow 3$, $1/\lambda_1 = R(\frac{1}{9} - \frac{1}{16}) = \frac{7R}{144}$. For $3 \rightarrow 2$, $1/\lambda_2 = R(\frac{1}{4} - \frac{1}{9}) = \frac{5R}{36}$. Ratio $\lambda_1/\lambda_2 = (144/7R) / (36/5R) = 20/7$.
Question 114:
easy20. Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr’s atomic model? (2011 Mains)
Energy levels of H-atom are $-13.6 eV$, $-3.4 eV$, $-1.51 eV$, $-0.85 eV$, etc. Possible photon energies are differences between these: $E_3-E_2 = 1.89 \approx 1.9 eV$, $E_4-E_3 = 0.66 \approx 0.65 eV$, $E_{\infty}-E_1 = 13.6 eV$. There is no transition corresponding to $11.1 eV$.
Question 115:
easyThe wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is:
(2011 Pre)
For first line of Lyman for H-atom ($n=2 \rightarrow 1$), $1/\lambda = R(1/1^2 - 1/2^2) = 3R/4$. For second line of Balmer for ion ($n=4 \rightarrow 2$), $1/\lambda = Z^2 R(1/2^2 - 1/4^2) = Z^2 R(3/16)$. Equating them gives $3R/4 = Z^2 R(3/16) \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.
Question 116:
easyThe electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be: (the energy of the electron in $n^{th}$ state)
(2010 Mains)
Energy of emitted photon $E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 eV$. Using Einstein's photoelectric equation, max kinetic energy $K_{max} = E - \phi = 12.09 - 5.1 = 6.99 eV \approx 7 eV$. Hence, the stopping potential is $7 V$.
Question 117:
easyThe energy of a hydrogen atom in the ground state is -13.6 eV. The energy of a $He^+$ ion in the first excited state will be:
(2010 Pre)
The energy of an electron in a hydrogen-like ion is $E_n = -13.6 \frac{Z^2}{n^2} eV$. For a $He^+$ ion, $Z=2$. The first excited state corresponds to $n=2$. Thus, $E_2 = -13.6 \frac{2^2}{2^2} = -13.6 eV$.
Question 118:
easyAccording to Bohr’s principle, the relation between principal quantum number (n) and radius of orbit (r) is
(1996)
The radius of the nth Bohr orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$. This shows that the radius is directly proportional to the square of the principal quantum number, so $r \propto n^2$.
Question 119:
easyAn electron makes a transition from orbit $n = 4$ to the orbit $n = 2$ of a hydrogen atom. What is the wavelength of the emitted radiations? (R – Rydberg’s constant)
(1995)
Using Rydberg's formula $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, we substitute $n_1=2$ and $n_2=4$. $\frac{1}{\lambda} = R \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$. This gives $\lambda = \frac{16}{3R}$.
Question 120:
easyWhen a hydrogen atom is raised from the ground state to an excited state,
(1995)
Kinetic energy $K.E. \propto \frac{1}{n^2}$, so it decreases as $n$ increases. Total energy $E = -\frac{13.6}{n^2} eV$ increases (becomes less negative). Since $P.E. = 2E$, potential energy also becomes less negative, meaning it increases.