Embibe Experts Solutions for Chapter: Dual Nature of Matter and Radiation, Exercise 3: Level 3

An electron accelerated through a potential of $10000 \mathrm{V}$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential, so that the wavelength is doubled?

The minimum intensity of light that can be perceived by human eye is about ${10}^{-10} \mathrm{W} {\mathrm{m}}^{-2}$. The number of photons of wavelength $5.6\times {10}^{-7} \mathrm{m}$ that must enter the pupil of area ${10}^{-6} {\mathrm{m}}^2$ per second for vision is approximately (use $h=6.6\times {10}^{-34} \mathrm{J} \mathrm{s}$)

is the wavelength of photon of energy $35\mathrm{keV}$ .
 

The correct curve between intensity of scattered electrons ($\mathrm{I}$) and the angle of diffraction ($\mathrm{\varphi }$) in Davisson-Germer experiment is:

The de Broglie wavlength associated with an electron having kinetic energy of $10 \mathrm{eV}$ is:

The energy that should be added to an electron to reduce its de Broglie wavelength from $1$ $\mathrm{nm}$ to $0.5 \mathrm{nm}$ is:

If the average kinetic energy of matter waves is $kT$, then the wavelength of de Broglie waves associated with neutrons at room temperature $\mathrm{T}$ is:

The work functions of tungsten and sodium are $5.06 \mathrm{eV}$ and $2.53 \mathrm{eV}$, respectively. If the threshold wavelength for sodium is $5896 \stackrel{\circ }{\mathrm{A}}$, then the threshold wavelength for tungsten will be