Dual Nature of ...

The figure shows the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. The ratio h/e is

Representing the stopping potential V along y-axis and $$(1/\lambda)$$ along x-axis for a given photocathode, the curve is a straight line, the slope of which is equal to

An electron of mass $$m_e$$ and a proton of mass $$m_p$$ are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an electron to that associated with proton is

The kinetic energy of a particle is equal to the energy of a photon. The particle moves at 5% of the speed of light. The ratio of the photon wavelength to the de Broglie wavelength of the particle is
[No need to used relativistic formula for the particle.]

For photoelectric effect in a metal, the graph of the stopping potential $$V_0$$ (in volt) versus frequency $$v$$ (in hertz) of the incident radiation is shown in Fig. The work function of the metal (in $$eV$$) is

A photon has same wavelength as the de Broglie wavelength of electrons. Given $$C=$$ speed of light, $$v=$$ speed of electron. Which of the following relation is correct? [Here $$E_e=$$ kinetic energy of electron, $$E_{ph}=$$ energy of photon, $$P_e=$$ momentum of electron and $$P_{ph}=$$ momentum of photon]

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

Two identical non-relativistic particles A and B move at right angles to each other, processing de Broglie wavelengths $$\lambda_1$$ and $$\lambda_2$$ respectively. The de Broglie wavelength of each particle in their centre of mass frame of reference is :

Two electrons are moving with same speed $$v$$. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, when after some time de Broglie wavelengths of two are $$\lambda_1$$ and $$\lambda_2$$, respectively. Now :

The magnitude of the de-Broglie wavelength $$(\lambda)$$ of electron (e), proton (p), neutron (n) and $$\alpha$$-particle $$(\alpha)$$ all having the same energy of 1MeV, in the increasing order will follow the sequence