Question

An electron of energy 150 eV has wavelength of 10^-10 m.the wavelength of 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.

According to de-Broglie hypothesis, 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,

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

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,

$E_1 = \dfrac{h^2}{2m\lambda_1^2}\\(150\ eV) = \dfrac{h^2}{2m(10^{-10})}\ \ .............\ (1).$

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

Therefore,

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

On dividing equations (1) and (2), we get,

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

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