Question 261:
easyIf we consider electrons and photons of same wavelength, then they will have same
(1995)
According to the de-Broglie relation, $\lambda = \frac{h}{p}$. If the wavelength $\lambda$ is the same, then the momentum $p$ must also be the same.
Question 262:
easyMomentum of photon wavelength $\lambda$ is
(1993)
The momentum of a photon is given by $p = \frac{h}{\lambda}$. Since $c = \nu \lambda$, we can write $\lambda = \frac{c}{\nu}$. Substituting this into the momentum equation gives $p = \frac{h\nu}{c}$.
Question 263:
easyThe wavelength of a $1 keV$ photon is $1.24 \times 10^{-9} m$. What is the frequency of $1 MeV$ photon?
(1991)
Energy $E = 1 MeV = 10^6 eV$. In joules, $E = 10^6 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-13} J$. Frequency $\nu = \frac{E}{h} = \frac{1.6 \times 10^{-13}}{6.63 \times 10^{-34}} \approx 2.4 \times 10^{20} Hz$.
Question 264:
easyThe momentum of a photon of an electromagnetic radiation is $3.3 \times 10^{-29} kg ms^{-1}$. What is the frequency of the associated waves? $[h = 6.6 \times 10^{-34} Js ; c = 3 \times 10^8 ms^{-1}]$
(1990)
Momentum $p = \frac{h\nu}{c}$. Frequency $\nu = \frac{pc}{h} = \frac{3.3 \times 10^{-29} \times 3 \times 10^8}{6.6 \times 10^{-34}} = \frac{9.9 \times 10^{-21}}{6.6 \times 10^{-34}} = 1.5 \times 10^{13} Hz$.
Question 265:
easyThe energy of a photon of wavelength $\lambda$ is
(1988)
The energy of a photon in terms of wavelength is given by the relation $E = \frac{hc}{\lambda}$.
Question 266:
easyThe wave nature of electrons was experimentally verified by.
(2020-Covid)
The wave nature of moving electrons was first experimentally verified by C.J. Davisson and L.H. Germer in 1927 through their electron diffraction experiment.
Question 267:
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 268:
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 269:
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 270:
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 $\geq 2.8 \times 10^{-9} m$.