Dual Nature Questions

A photon and an electron (mass $m$) have the same energy $E$. The ratio $({\lambda }_{\text{photon}}/{\lambda }_{\text{electron}})$ of their de-Broglie wavelengths is ($c$ is the speed of light):

Solid targets of different elements are bombarded by highly energetic electron beams. The frequency ($f$) of the characteristic X-rays emitted from different targets varies with atomic number $Z$ as:

Assertion: The energy ($E$) and momentum ($p$) of a photon are related by $p=E/c$.Reason: The photon behaves like a particle.

Assertion: For higher temperatures the peak emission wavelength of a black body shifts to lower wavelengths.Reason: Peak emission wavelengths of a blackbody is proportional to the fourth-power of temperature.

A particle of mass $1\text{mg}$ has the same wavelength as an electron moving with a velocity of $3\times 10^6{\text{ms}}^{-1}$. The velocity of the particle is ($mass$ $of$ $electron=9.1\times 10^{-31}\text{kg}$):

Monochromatic light of wavelength $667\text{nm}$ is produced by a helium neon laser. The power emitted is $9\text{mW}$. The number of photons arriving per sec. on the average at a target irradiated by this beam is:

If the momentum of an electron is changed by $P$, then the de Broglie wavelength changes by $0.5\%$. The initial momentum of electron was:

An $\alpha$ -particle moves in a circular path of radius $0.83\text{cm}$ in the presence of a magnetic field of $0.25{\text{Wb/m}}^2$. The de Broglie wavelength associated with the particle will be:

An electron of mass $m$ and a photon have same energy $E$. The ratio of de-Broglie wavelengths associated with them is:

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength (${\lambda }_0$) of the emitted X-ray is: