Calculate the wavelength of an electron that has been accelerated in a

1.	Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potential difference of 1kev
Answer» Answer: Explanation: 1/2 mv²=ev------------1 e= CHARGE on electron=1.6 x 10⁻¹⁹ Coulombs V=1000v mass of electron=m=9.1 x10⁻³¹ kg Substituting these values in EQUATION 1, we get 1/2 x9.1 x10⁻³¹ x v² = 1.6 x 10⁻¹⁹ x1000 v²=2x1.6 x 10⁻¹⁹ x1000/9.1 x10⁻³¹ v²=3.2x10⁻¹⁹x1000/9.1 x10⁻³¹ v²=3.5x101⁴ v=√3.5x10⁴=1.88 x 10⁷ m/s de-Broglie WAVELENGTH of an electron = λ=h/mv where h= 6.67 x 10 ⁻³⁴ J-s λ= 6.67 x 10 ⁻³⁴ /9.1 x10⁻³¹ x1.88 x 10⁷ =3.87 x10⁻¹¹ m ∴de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV is 3.87 x10⁻¹¹ m #sneha Plz support me MARK it as brainalist

Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potential difference of 1kev

Answer»

Answer:

Explanation:

1/2 mv²=ev------------1

e= CHARGE on electron=1.6 x 10⁻¹⁹ Coulombs

V=1000v

mass of electron=m=9.1 x10⁻³¹ kg

Substituting these values in EQUATION 1, we get

1/2 x9.1 x10⁻³¹ x v² = 1.6 x 10⁻¹⁹ x1000

v²=2x1.6 x 10⁻¹⁹ x1000/9.1 x10⁻³¹

v²=3.2x10⁻¹⁹x1000/9.1 x10⁻³¹

v²=3.5x101⁴

v=√3.5x10⁴=1.88 x 10⁷ m/s

de-Broglie WAVELENGTH of an electron = λ=h/mv

where h= 6.67 x 10 ⁻³⁴ J-s

λ= 6.67 x 10 ⁻³⁴ /9.1 x10⁻³¹ x1.88 x 10⁷

=3.87 x10⁻¹¹ m

∴de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 keV is 3.87 x10⁻¹¹ m

#sneha

Plz support me MARK it as brainalist