Question 51:
easyThe total energy of an electron in the first excited state of hydrogen is about -3.4 eV. Its kinetic energy in this state is:
(2005)
For an electron in an orbit, its kinetic energy is the negative of its total energy ($K = -E$). Given the total energy is $-3.4 eV$, its kinetic energy is $-(-3.4 eV) = 3.4 eV$.
Question 52:
easyIn the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
(2015 Pre)
For longest wavelength in Lyman series ($n_2=2$ to $n_1=1$), $1/\lambda_L = R(1 - 1/4) = 3R/4$. For longest wavelength in Balmer series ($n_2=3$ to $n_1=2$), $1/\lambda_B = R(1/4 - 1/9) = 5R/36$. Ratio $\lambda_L/\lambda_B = (4/3R) / (36/5R) = 20/108 = 5/27$.
Question 53:
easyEnergy levels A, B and C of a certain atom correspond to increasing values of energy i.e., $E_A < E_B < E_C$. If $\lambda_1, \lambda_2$ and $\lambda_3$ are wavelengths of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct?
(2005)
From energy conservation, $E_C - E_A = (E_C - E_B) + (E_B - E_A)$. Using $E = \frac{hc}{\lambda}$, we have $\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$. Dividing by $hc$ gives $\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$, which rearranges to $\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.
Question 54:
easyHydrogen atom in ground state is excited by a monochromatic radiation of $\lambda = 975 A$. Number of spectral lines in the resulting spectrum emitted will be:
(2014)
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{12400}{975} eV \approx 12.75 eV$. The atom reaches an energy of $-13.6 + 12.75 = -0.85 eV$, which corresponds to $n = 4$ state since $-13.6/4^2 = -0.85 eV$. Number of spectral lines is $n(n-1)/2 = 4(3)/2 = 6$.
Question 55:
easyRatio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
(2013)
Longest wavelength in Lyman is $\lambda_L = \frac{4}{3R}$. Longest wavelength in Balmer is $\lambda_B = \frac{36}{5R}$. Their ratio is $\lambda_L / \lambda_B = \frac{4}{3R} \times \frac{5R}{36} = \frac{20}{108} = \frac{5}{27}$.
Question 56:
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)
Ultraviolet radiation corresponds to Lyman series (transition to $n=1$). Infrared radiation corresponds to Paschen (to $n=3$), Brackett (to $n=4$), etc. Among the options, $4 \rightarrow 3$ belongs to the Paschen series, which emits infrared radiation.
Question 57:
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 58:
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 59:
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 60:
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 $11.1 eV$.