Dual Nature of ...

The de-Broglie's wave length for an electron in $$10$$th Bohr's orbit  H-atom is [$$ h= 6.625 \times 10^{-34}$$ j/s, $$m_e = 9.1 \times 10^{-31}$$ kg,and $$a_o =0.529$$ angstrom] 

Energy required to stop the ejection of electrons from $$Cu$$ plate is $$0.24ev$$. Calculate the work function when radiations of wavelength $$254nm$$ strikes the plate

Consider three election jumps described below for the hydrogen atom $$X:n=3$$ to $$n=1$$           $$Y:n=4$$ to $$n=2$$             $$Z:n=5$$ to $$n=3$$
The proton emitted in which transition $$X,Y$$(or)$$Z$$ will be have shortest wavelength

If the shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is $$0.1\mathring{A}$$, then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately 
($$hc=12400eV\mathring{A}$$, $$h=6.63\times{10}^{-34}$$ in MKS, mass of electron $$=9.1\times{10}^{-31}kg$$)

A particle of mass $$M$$ at rest decays into two particles of masses $$m _ { 1 } \text { and } m _ { 2 }$$ having non zero velocities. The ratio of the de-Broglie wavelengths of the particles $$\lambda _ { 1 } / \lambda _ { 2 }$$ is

The de Broglie wavelengths associated with a proton and an $$\alpha$$- particle will be:

A hydrogen atom moving at a speed $$V$$ absorbs a photon of wavelength $$103nm$$ and stops. Now the hydrogen atom starts moving in the backward direction with velocity $${V}^{1}$$. The possible value of $${V}^{1}$$ is (approximately)

If given particles are moving with same velocity, then maximum de-Broglie wavelength for

An $$X$$- rays tube operate at $$10 KV$$. The ratio of X- rays wavelength to that of De - Broglie is    

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $$0.5\ V$$ When the radiation of $$250\ nm$$  is used. The work function of the metal is: