Question 41:
easyLet $T_1$ and $T_2$ be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to the Bohr’s model of an atom, the ratio $T_1 : T_2$ is:
(2022)
Energy in Bohr's model is $E_n \propto \frac{1}{n^2}$. The first excited state is $n=2$, so $T_1 \propto \frac{1}{4}$. The second excited state is $n=3$, so $T_2 \propto \frac{1}{9}$. The ratio $T_1 : T_2 = \frac{1}{4} : \frac{1}{9} = 9:4$.
Question 42:
easyFor which one of the following, Bohr’s model is not valid?
(2020)
Bohr's model is only applicable to single-electron species (hydrogen-like atoms). Singly ionised neon ($Ne^+$) has 9 electrons, so Bohr's model is not valid for it.
Question 43:
easyThe total energy of an electron in the $n^{th}$ stationary orbit of the hydrogen atom can be obtained by.
(2020-Covid)
According to Bohr's theory, the total energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} eV$.
Question 44:
easyGiven the value of Rydberg constant is $10^7 m^{-1}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be:
(2016 – I)
The wave number $\bar{\nu}$ is $1/\lambda$. For the last line of the Balmer series, $n_1 = 2$ and $n_2 = \infty$. Thus, $\bar{\nu} = R(\frac{1}{2^2} - 0) = \frac{R}{4} = \frac{10^7}{4} = 0.25 \times 10^7 m^{-1}$.
Question 45:
easyThe total energy of an electron in an atom in an orbit is -3.4 eV. Its kinetic and potential energies are, respectively:
(2019)
For an electron in an orbit, kinetic energy is equal to the negative of total energy: $K = -E = -(-3.4 eV) = 3.4 eV$. Potential energy is twice the total energy: $U = 2E = 2 \times (-3.4 eV) = -6.8 eV$.
Question 46:
easyThe ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
(2018)
The relationship between kinetic energy $K$, potential energy $U$, and total energy $E$ for an electron in a Bohr orbit is $K = -E$ and $U = 2E$. Therefore, the ratio of kinetic energy to total energy is $1 : -1$.
Question 47:
easyThe ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
(2017-Delhi)
The last line of a series corresponds to $n_2 = \infty$. For the Balmer series, $1/\lambda_B = R(\frac{1}{2^2} - 0) \Rightarrow \lambda_B = \frac{4}{R}$. For the Lyman series, $1/\lambda_L = R(\frac{1}{1^2} - 0) \Rightarrow \lambda_L = \frac{1}{R}$. The ratio is $\lambda_B / \lambda_L = 4$.
Question 48:
easyIf an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:
(2016 – II)
For the first transition, $1/\lambda = R(\frac{1}{2^2} - \frac{1}{3^2}) = \frac{5R}{36}$. For the second transition, $1/\lambda' = R(\frac{1}{3^2} - \frac{1}{4^2}) = \frac{7R}{144}$. Dividing the two gives $\lambda' = \lambda \times \frac{144/7}{36/5} = \frac{20}{7}\lambda$.
Question 49:
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 50:
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, $v_n = 2.18 \times 10^6 \times \frac{2}{3} = 1.453 \times 10^6 m/s \approx 1.46 \times 10^6 m/s$.