Question

The de broglie eavelength of a photon is twice that of electron.The speed of electron is c/100.The
Ee/Ep=10^(-4)
Ee/Ep =10^(-2)

Answer 1

Correct Question

The De Beogile wavelength of a photon is twice that of an electron. The speed of electron is c/100. The ratio of their energies would be :

Solution

Given :

• The wavelength of the photon is twice that of an electron

• The speed of electron is c/100

Assume the wavelength of the electron to be $\lambda$. Implies,the wavelength of the photon would be $\sf 2 \lambda$

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We know that,

$\huge{ \boxed{ \boxed{ \sf{E = h \: \nu}}}}$

$\sf{Here}\begin{cases}\sf{E \longrightarrow Energy} \\ \sf{h \longrightarrow Planck's \ constant} \\ \sf{\nu \longrightarrow Frequency }\end{cases}$

Also,

$\large{ \boxed{ \boxed{ \sf{ \nu \: = \dfrac{c}{ \lambda} }}}}$

$\sf{Here} \begin{cases} \sf{c \longrightarrow Speed \ of \ Light} \\ \sf{\lambda \longrightarrow Wavelength} \end{cases}$

Implies

$\longmapsto \: \boxed{ \boxed{\sf \: E = \dfrac{hc}{ \lambda} }}$

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For the photon,

$\tt{E_p = \dfrac{hc}{2 \lambda} - - - - - - - (1)}$

For the electron,

$\tt{E_e = \dfrac{hc}{100 \lambda} - - - - - - - - (2) }$

Dividing equations (1) and (2),we get :

$\implies \: \sf{ \dfrac{E_p}{E_e} = \dfrac{hc}{2 \lambda} \times \dfrac{100 \lambda}{hc} } \\ \\ \implies \: \underline{ \boxed{ \sf{E_p \: \colon \: E_e = 50 \: \colon \: 1}}}$

The ratio of their energies is 50 : 1

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