Dual Nature of ...

If the stationery proton and $$\alpha-$$ particle are accelerated through same potential difference, the ratio of de-Broglie wavelength will be:

The de-Broglie wavelength of an electron in first orbit of Bohr hydrogen is equal to

If $$E_{1},E_{2}$$ and $$E_{3}$$ represent respectively the kinetic energies of an electron, an $$\alpha-$$particle and a proton each having same de-Broglies wavelength then

If the kinetic energy of a free electron doubles, its de Broglie wavelength changes by the factor

After absorbing a slowly moving neutron of mass $${m}_{N}$$ (momentum $$\sim  0$$) a nucleus of mass $$M$$ breaks into two nuclei of masses $${m}_{1}$$ and $$5{ m }_{ 1 }\left( 6{ m }_{ 1 }=M+{ m }_{ N } \right) $$ respectively. If the de-Broglie wavelength of the nucleus with mass $${m}_{1}$$ is $$\lambda$$, the de-Broglie wavelength of the other nucleus will be:

Photon of frequency $$v$$ has a momentum associated with it. If $$c$$ is the velocity of light, the momentum is :

The graph shown below depicts plot of photocurrent versus anode potential for a cathode with 4 eV work function. The energy of the incident photon is

Let $$p$$ and $$E$$ denote the linear momentum and the energy of a photon. For another photon of smaller wavelength (in same medium)

A proton and electron are accelerated by same potential difference starting from the rest have de-Broglie wavelength $$\lambda_p$$ and $$\lambda_e$$.

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let $${ \lambda  }_{ 1 }$$ be de-Broglie wavelength of the proton and $${ \lambda  }_{ 2 }$$ be the wavelength of the photon.The ration $$\dfrac { { \lambda  }_{ 1 } }{ { \lambda  }_{ 2 } }$$ is proportional to: