Dual Nature of ...

The radius of the second orbit of an electron in hydrogen atom is $$2.116\overset {o}{A}$$. The de Broglie wavelength associated with this electron in this orbit would be

A particle of mass 3m at rest decays into two particles of masses m and 2m having non-zero velocities. The ratio of the de Broglie wavelengths of the particles $$(\lambda_1/\lambda_2)$$ is

The de Broglie wavelength of a thermal neutron at $$927^oC$$ is $$\lambda$$. Its wavelength at $$327^oC$$ will be

The energy of a photon is equal to the kinetic energy of a proton. The energy of photon is $$E$$. Let $$\lambda_1$$ be the de Broglie wavelength of the proton and $$\lambda_2$$ be the wavelength of the photon. Then, $$\lambda_1/\lambda_2$$ is proportional to :

Einstein's Photo electric equation:

We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used the energy required should be

Calculate the de Broglie wavelength of an electron having kinetic energy of $$1.6\times 10^{-6}erg$$ $$(m_e=9.11\times 10^{-28}g, h=6.62\times 10^{-27}erg-sec)$$.

The wavelength of a wave is $$\lambda$$ = 6000 $$\overset{o}{A}$$, then wave number will be

Einstein's photoelectric equation state that