Question 81:
easyIn the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by
(2011 Pre)
The velocity of electrons is determined by the accelerating voltage. Therefore, increasing the potential difference between the anode and the filament increases the kinetic energy and hence the velocity of the emitted electrons.
Question 82:
easyThe interplanar distance in a crystal is $2.8 \times 10^{-8} m$. The value of maximum wavelength which can be diffracted:
(2001)
According to Bragg's law, $2d \sin \theta = n\lambda$. For maximum wavelength, $\sin \theta$ must be maximum ($=1$) and order $n = 1$. So $\lambda_{max} = 2d = 2 \times 2.8 \times 10^{-8} m = 5.6 \times 10^{-8} m$.
Question 83:
easyThe de Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it
(1989)
The de Broglie wavelength of a particle of mass m and velocity v is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Question 84:
easyLight of wavelength $5000 nm$ is incident on a metal with work function $2.28 eV$. The de-Broglie wavelength of the emitted electron is:
(2015 Re)
Energy of incident light (assuming it was meant to be $500 nm$ for standard photoelectric emission since $5000 nm$ is only $0.248 eV$) $E = \frac{1240}{500} = 2.48 eV$. Maximum kinetic energy of emitted electron $K_{max} = E - \phi = 2.48 - 2.28 = 0.2 eV$. The minimum de-Broglie wavelength $\lambda_{min} = \frac{12.27}{\sqrt{0.2}} \AA \approx 27.4 \AA = 2.74 \times 10^{-9} m$. Therefore, the wavelength is $\geq 2.8 \times 10^{-9} m$.
Question 85:
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 86:
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 87:
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 88:
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 89:
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 90:
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 $r_0 \propto \frac{1}{m}$.