Calculate the wavelength of an electron in a 10 MeV particle accelerator (1 MeV = 10⁶ eV).
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Given conditions ⇒
Energy of the Electron = 10 MeV
= 10 × 10⁶ eV.
= 10⁷ × 1.6 × 10⁻¹⁹ J.
= 16 × 10⁻¹³ J.
Now, We know the Mass of the Electron = 9.1 × 10⁻³¹ kg.
∵ E = 1/2 × mv²
where, E is the Energy, m is the mass of the electrons, and v is the velocity of the Electrons.
∴ 16 × 10⁻¹³ = 1/2 × 9.1 × 10⁻³¹ × v²
∴ v² = (32 × 10³¹ × 10⁻¹³)÷9.1
⇒ v² = 3.51 × 10¹⁸
⇒ v = √(3.51 × 10¹⁸)
⇒ v = 1.86 × 10⁹ m/s.
Momentum(p) = m × v
= 9.1 × 10⁻³¹ × 1.86 × 10⁹
= 16.926 × 10⁻²²
Now, Using the Formula,
λ = h/p
where h is the plank's constant = 6.63 × 10⁻³⁴ J-s.
⇒ λ = (6.63 × 10⁻³⁴) ÷ (16.926 × 10⁻²²)
⇒ λ = 0.39 × 10⁻¹² m.
∴ λ = 0.39 pm. [∵ 1 pm = 10⁻¹².]
Hence, the de-broglie wavelength of electron is 0.39 pm.
Hope it helps.
Energy of the Electron = 10 MeV
= 10 × 10⁶ eV.
= 10⁷ × 1.6 × 10⁻¹⁹ J.
= 16 × 10⁻¹³ J.
Now, We know the Mass of the Electron = 9.1 × 10⁻³¹ kg.
∵ E = 1/2 × mv²
where, E is the Energy, m is the mass of the electrons, and v is the velocity of the Electrons.
∴ 16 × 10⁻¹³ = 1/2 × 9.1 × 10⁻³¹ × v²
∴ v² = (32 × 10³¹ × 10⁻¹³)÷9.1
⇒ v² = 3.51 × 10¹⁸
⇒ v = √(3.51 × 10¹⁸)
⇒ v = 1.86 × 10⁹ m/s.
Momentum(p) = m × v
= 9.1 × 10⁻³¹ × 1.86 × 10⁹
= 16.926 × 10⁻²²
Now, Using the Formula,
λ = h/p
where h is the plank's constant = 6.63 × 10⁻³⁴ J-s.
⇒ λ = (6.63 × 10⁻³⁴) ÷ (16.926 × 10⁻²²)
⇒ λ = 0.39 × 10⁻¹² m.
∴ λ = 0.39 pm. [∵ 1 pm = 10⁻¹².]
Hence, the de-broglie wavelength of electron is 0.39 pm.
Hope it helps.
abhi178:
Bhiaya check first step , 10 MeV = 1.6 × 10^-12 J
Answered by
2
Since rest mass energy of electron is 0.51 MeV so it has to be a reltivistic motion and we can't use the formulae P^2 /2m = E E^2 = p^2*c^2 + mc^2 will pe used to calculate p
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