Dual Nature of ...

An electron is continuously accelerated in vacuum tube by applying potential difference.  If  its de Broglie wavelenght is decreased by 1.0 % the change in the kinetic energy of ht eelectron is nearly : 

Energy of an electron is given by
$$E=-2.178\times 1{ 0 }^{ -18 }J\left( \dfrac { { Z }^{ 2 } }{ { n }^{ 2 } }  \right) $$.
Wavelength of light required to excite an electron in an hydrogen atom from level n=1 to n=2 will be
$$(h=6.62\times 1{ 0 }^{ -34 }J.s\quad and\quad c=3.0\times { 10 }^{ 8 }m/s)$$

When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x=0 and x=a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as E =$$\frac{\rho^2}{2m}$$. Thus, the energy of the particle can be denoted by a quantum number n taking values 1, 2, 3, . (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = $$6.6 \times 10^{-34} \ Js $$ and $$ e =1.6 \times 10^{-19} $$ C.
The speed of the particle that can take discrete values is proportional to

A stream o electons from a heated filament was passed between two charged plates kept at a potential difference V. If e and m are charge and mass of an electron respectively,them the value of $${ h }/{ \lambda  }$$ (where$$ { \lambda  }$$ is wavelength associated with electon wave) is gien by:

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 form level  $$n=1$$ to n = 2 will be:

The wavelength of radiation emitted, WHEN in $${ He }^{ + }$$ electron falls from infinity to stationary state would be $$(R=1.097\times \quad { 10 }^{ 7 }m^{ -1 })$$

An object in space is moving with a velocity of $$10^{10}cm/s$$ towards a star that emits radiations of wavelength $$6\times 10^{-5}cm$$. The wavelength of the radiations recived by the crew in the object will be:

Two particles A and B are moving in same direction have deBroglie wavelengths $${ \lambda  }_{ 1 }$$ and $${ \lambda  }_{ 2 }$$ combine to from a particle C. The process conserve momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional)

Calculate the wavelength (in nanometre) associated with a proton moving at $$ 1 \times {10}^{ 3 }m{ s }^{ -1 }$$.

A certain dye absorbs light of certain wavelength and then fluorescence light of wavelength $${\text{5000}}\mathop {\text{A}}\limits^{\text{o}} $$Assuming that under given conditions, 50% of the absorbed energy is re-emitted out as flourescence and thr ratio of number of quanta emitted out to the number of quanta absorbed is 5 : 8, Find the wavelength of absorbed $$\;\;\left( {{\text{in}}\;\mathop {\text{A}}\limits^{\text{o}} } \right)\;:\;\left[ {{\text{hc}} = {\text{12400}}\;{\text{eV}}\mathop {\text{A}}\limits^{\text{o}} } \right]\;\;\;\;$$