Question 101:
easyThe ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is:
(2008)
Excitation energy is the energy required to excite the electron from the ground state ($n=1$) to a particular state. The energy of the first excited state ($n=2$) is $-13.6/4 = -3.4 eV$. The excitation energy is $-3.4 - (-13.6) = 10.2 eV$.
Question 102:
easyThe total energy of electron in the ground state of hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is:
(2007)
The total energy in the first excited state ($n=2$) is $E_2 = -13.6/2^2 = -3.4 eV$. Since kinetic energy $K = -E_n$, the kinetic energy in the first excited state is $3.4 eV$.
Question 103:
easyIonisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be:
(2006)
Initial energy is $-13.6 eV$. After absorbing $12.1 eV$, the final energy is $-13.6 + 12.1 = -1.5 eV$. This corresponds to the $n=3$ state (since $-13.6/3^2 \approx -1.51 eV$). Number of spectral lines emitted upon returning to ground state is $\frac{3(3-1)}{2} = 3$.
Question 104:
easyConsider 3rd orbit of $He^+$ (Helium) using non relativistic approach the speed of electron in this orbit will be (given $K = 9 \times 10^9$ constant $Z = 2$ and h (Planck’s constant) = $6.6 \times 10^{-34} Js$):
(2015)
Speed of electron in nth orbit is $v_n = 2.18 \times 10^6 \frac{Z}{n} m/s$. For $He^+$, $Z=2$ and $n=3$. So, $v_n = 2.18 \times 10^6 \times \frac{2}{3} = 1.453 \times 10^6 m/s \approx 1.46 \times 10^6 m/s$.
Question 105:
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 106:
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 107:
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 108:
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 109:
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 110:
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.