Dual Nature of ...

Protons are accelerated from rest by a potential difference 4 kV and strike a metal target. If a proton produces one photon on impact of minimum wavelength $$\lambda_1$$ and similarly an electron accelerated to 4 kV strikes the target and produces a minimum wavelength $$\lambda_2$$ then

In the following diagram if $$\displaystyle V_{2}>V_{1}$$then

The kinetic energy of photoelectrons depends on 

An electron is confined in the region of width 1 $$ A^{\circ} $$. Estimate its K.E.

Neglecting variation of mass with energy the wavelength associated with an electron having a kinetic energy $$E$$ is proportional to 

If $$\lambda_1$$ and $$\lambda_2$$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom, then $$\lambda_1/\lambda_2$$ is equal to :

A particle of mass M at rest decays into two masses $$m_1$$ and $$m_2$$ with non-zero velocities. The ratio $$\lambda_1 / \lambda_2$$ of de Broglie wavelengths of particles is

A particle of mass m is projected from ground with velocity u making angle $$\theta$$ with the vertical. The de Broglie wavelength of the particle at the highest point is

Two electrons are moving with the same velocity, one electron enters a region of uniform electric field while the other enters a region of uniform magnetic field. Then, after sometime, if the de-Broglie wavelength of two are $$\lambda_{1}$$ and $$\lambda_{2}$$ respectively, then

The energy $$E$$ and the momentum $$p$$ of a photon is given by $$E = hv$$ and $$p = \displaystyle\ \frac{h}{\lambda}$$. The velocity of photon will be