Question 271:
easyIf the momentum of an electron is changed by $P$, then the de-Broglie wavelength associated with it changes by $0.5\%$. The initial momentum of electron will be:
(2012 Mains)
de-Broglie wavelength $\lambda = \frac{h}{p}$. Differentiating, $|\frac{\Delta \lambda}{\lambda}| = \frac{\Delta p}{p}$. Given $|\frac{\Delta \lambda}{\lambda}| = 0.005$ and $\Delta p = P$. So, $0.005 = \frac{P}{p_{initial}} \implies p_{initial} = \frac{P}{0.005} = 200 P$.
Question 272:
easyThe momentum of a photon of energy $1 MeV$ in $kg m/s$ will be
(2006)
Momentum of a photon is $p = \frac{E}{c}$. Energy $E = 1 MeV = 10^6 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-13} J$. $p = \frac{1.6 \times 10^{-13}}{3 \times 10^8} \approx 5.33 \times 10^{-22} kg m/s$.
Question 273:
easyThe value of Planck’s constant is:
(2002)
Planck's constant has the value $6.63 \times 10^{-34} J \cdot s$. Since $1 J = 1 kg \cdot m^2/s^2$, the unit $J \cdot s$ is equivalent to $kg-m^2/s$.
Question 274:
easy84. In a discharge tube at $0.02 mm$, there is formation of (1996)
At a very low pressure of about $0.02 mm$ of Hg in a discharge tube, the Crookes dark space expands to fill the entire tube.
Question 275:
easyWhen an $\alpha$ particle of mass m moving with velocity v bombards on a heavy nucleus of charge ‘Ze’, its distance of closest approach from the nucleus depends on mass:
(2016 – I)
At the distance of closest approach $r_0$, kinetic energy is converted to potential energy. $\frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$. Rearranging gives $r_0 = \frac{4 Z e^2}{4\pi\epsilon_0 m v^2}$, which shows $r_0 \propto \frac{1}{m}$.
Question 276:
easyAn alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to:
(2010 Pre)
The distance of closest approach is found by equating kinetic energy to electrostatic potential energy: $K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$. Since $K = \frac{1}{2} m v^2$, we have $r_0 \propto \frac{1}{m}$.
Question 277:
easyIn 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 $z_1 z_2$.
Question 278:
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 279:
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 280:
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 $E_n = -\frac{13.6}{n^2} eV$.