The de Broglie wavelength (\lambda) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency] :

  • Option 1)

    \lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{1}{4}}}

     

     

     

  • Option 2)

    \lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{3}{2}}}

     

  • Option 3)

    \lambda \: \alpha \frac{1}{(v-v_{0})}

  • Option 4)

    \lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{1}{2}}}

Answers (1)
A admin

 

De-broglie wavelength -

\lambda = \frac{h}{mv}= \frac{h}{p}

- wherein

where m is the mass of the particle

v its velocity 

p its momentum

 

Orbital frequency -

f=\frac{mz^{2}e^{4}}{4\epsilon _{2}^{0}n^{3}h^{3}}

- wherein

f\alpha \frac{z^{2}}{n^{3}}

m & e are mass and charge of electron 

h = planck's constant

z = atomic number

n = principal quantum number

Formula 

hv = hv_{o}+\frac{1}{2}mv^{2}

h (v-v_{o})=\frac{1}{2}mv^{2}

v = (\frac{2h(v-v_{o})}{m})^{v_{2}}

\lambda =\frac{h}{mv}=\frac{h}{m}\frac{\sqrt{m}}{\sqrt{2h(v-v_{o})}}=\sqrt{\frac{h}{m(v-v_{o})}}

\lambda \alpha \frac{1}{(v-v_{o})^{v_{2}}}

???????

  


Option 1)

\lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{1}{4}}}

 

 

 

Option 2)

\lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{3}{2}}}

 

Option 3)

\lambda \: \alpha \frac{1}{(v-v_{0})}

Option 4)

\lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{1}{2}}}

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