Dual Nature of ...

Energy of an electron is given by $$E=-2.178\times { 10 }^{ -18 }J\left( \dfrac { { Z }^{ 2 } }{ n^{ 2 } }  \right) .$$ Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be ($$h=6.62\times { 10 }^{ -34 }Js$$ and $$c=3.0\times { 10 }^{ 8 }{ ms }^{ -1 }$$)

If the total energy of an electron in a hydrogen atom in excited state is -3,4 eV, then the de Broglie wavelength of the electron is :-

The uncertainty in location of circulating electron is equal to its de Broglie wave length. The minimum present error in its measurement of velocity under this circumstance will be approximately.

Calculate the wavelength (in nanometer) associated with a proton moving at $$1.0\times { 10 }^{ 3 }m{ s }^{ -1 }$$ 
(Mass of proton = $$1.67\times { 10 }^{ -27 }kg$$ and $$h=6.63\times { 10 }^{ -34 }Js$$)

The kinetic energy of one electron is $$2.8\times { 10 }^{ -13 }J.$$ What is the de-Broglie wavelength (Mass of electron $$9.1\times { 10 }^{ -13 }$$  kg)

Ionization Energy per mole of Na atom is $$  4.946 \times 10^{2} \mathrm{kJ/mole} $$ . The wavelength of radiation sufficient to ionize sodium atom is

Calculate kinetic energy of an electron emitted from a surface of potassium metal $$(work function=3.62\times{ 10 }^{ -16 }J)$$ by light of wavelength $$3\times10^{ -10 }$$m

If $$  \lambda_{0}  $$ is the threshold wavelength for photoelectric emission, $$\lambda$$ wavelength of light falling on the surface of the metal, and m is the mass of an electron, then de Broglie wavelength of the emitted electron is :

The energy required to break one mole of Cl-Cl bonds is $$242kJ/mol $$. The longest wavelength of light capable of breaking a single Cl-Cl bond is 

A gas absorbs a photon of 355 nm and emits at two wavelengths If one of the emission is at 680 nm, The other is at