Question

An electron of energy 150 eV has wavelength of 10^-10m. The wavelength of a 0.60 keV electron is??

Answer 1

The wavelength of 0.60 keV electron will be 0.5 x 10⁻¹⁰ m

Explanation:

We know that wavelength of an electron is given by the relation,

λ = $\frac{h}{\sqrt{2mV} }$

where,

h = planks constant

m = mass of electron

V = energy of the electron,

keeping h and m constant, we have,

λ ∝ 1/√V

=> λ₂/λ₁ = √(V₁/V₂)

=> λ₂ = 10⁻¹⁰ x √(150/600)

=> λ₂ = 10⁻¹⁰ x 1/2 = 0.5 x 10⁻¹⁰ m

Hence the wavelength of 0.60 keV electron will be 0.5 x 10⁻¹⁰ m

Answer 2

Answer:

$5.0\times 10^{-11}\ m$.

Explanation:

The energy of an electron is given by

$E=\dfrac{p^2}{2m}$,

where,

$p$ = momentum of the electron.

$m$ = mass of the electron.

The de-Broglie hypothesis states that the momentum of the electron in terms of its wavelength is given by

$p=\dfrac{h}{\lambda}$

where,

$h$ = Planck's constant.

$\lambda$ = wavelength of the electron.

Therefore, energy of the electron is given by

$E=\dfrac{h^2}{2m\lambda^2}$

Given that,

Energy of the electron is 150 eV when the wavelength of the electron is $10^{-10}\ m$.

Let,

$E_1=150\ eV\\\lambda_1=10^{-10}\ m$

Therefore,

$E_1=\dfrac{h^2}{2m\lambda_1^2}\\150\ eV=\dfrac{h^2}{2m(10^{-10})^2}\ \ .......\ (a).$

When the energy is $0.60\ keV$, say, $E_2=0.60\ keV$, let the wavelength of the electron corresponding to this energy be $\lambda_2$.

$E_2=\dfrac{h^2}{2m\lambda_2^2}\\0.60\ keV=\dfrac{h^2}{2m\lambda_2^2}\ \ .......\ (b).$

On dividing equations (a) and (b), we get,

$\dfrac{150\ eV}{0.60\ keV}=\dfrac{\dfrac{h^2}{2m(10^{-10})^2}}{\dfrac{h^2}{2m\lambda_2_2^2}}\\\dfrac{\lambda_2^2}{(10^{-10}\ m)^2}=\dfrac{150\ eV}{0.60\times 10^3\ eV}\\\lambda_2^2=(10^{-10})^2\times \dfrac{150\ eV}{0.60\times 10^3\ eV}=2.5\times 10^{-21}\\\lambda_2 = 5.0\times 10^{-11}\ m$

Thus, the required wavelength is $5.0\times 10^{-11}\ m$.