Modern Physics - NEET Physics Chapterwise MCQs & PYQs
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NEET Modern Physics MCQs & PYQs
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Question 91:
easy
In a Rutherford scattering experiment, when a projectile of charge $z_1$ and mass $M_1$ approaches a target nucleus of charge $z_2$ and mass $M_2$, the distance of closest approach is $r_0$. The energy of the projectile is
(2009)
At the distance of closest approach $r_0$, the entire kinetic energy of the projectile is converted into electrostatic potential energy. Energy $E = \frac{1}{4\pi\epsilon_0} \frac{z_1 z_2}{r_0}$. Thus, the energy is directly proportional to the product of charges $z_1 z_2$.
Let $T_1$ 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 $T_1 : T_2 = \frac{1}{4} : \frac{1}{9} = 9:4$.
For 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.
The 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 $E_n = -\frac{13.6}{n^2} eV$.
The 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 $1 : -1$.
The 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 $\lambda_B / \lambda_L = 4$.
If 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 $\lambda' = \lambda \times \frac{144/7}{36/5} = \frac{20}{7}\lambda$.
Given 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, $\bar{\nu} = R(\frac{1}{2^2} - 0) = \frac{R}{4} = \frac{10^7}{4} = 0.25 \times 10^7 m^{-1}$.
The 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: $U = 2E = 2 \times (-3.4 eV) = -6.8 eV$.
24. The ionisation energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between: (2009)
Number of spectral lines emitted is $\frac{n(n-1)}{2} = 6 \Rightarrow n = 4$. Maximum wavelength corresponds to the minimum energy difference. For transitions among levels up to $n=4$, the transition $4 \rightarrow 3$ has the minimum energy difference and thus the maximum wavelength.