The accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.05 nm is (wt. of one mole of proton =1.008 g )

Question

The accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.05 nm is (wt. of one mole of proton =1.008 g ) (A) 0.325 eV (B) 5.205 eV (C) 52.05 eV (D) 3.25 eV

Tardigrade Admin · Accepted Answer

Weight of one proton =(1.008 × 10-5/6.023 × 1023) kg We know, λ=(h/m v) v=(h/m λ), λ=0.05 nm =0.05 × 10-9 m =5 × 10-11 m K . E .=(1/2) m v2=(1/2) m((h/m λ))2 =(1/2) (h2/m λ2) =(1/2) × ((6.6 × 10-34)2 × 6.023 × 1023/1.008 × 10-3 ×(5 × 10-11)2) =5.205 × 10-20 J =(5.205 × 10-20/1.6 × 10-19) eV =3.25 × 10-1 =0.325 eV

Q. The accelerating potential that must be imparted to a proton beam to give it an effective wavelength of $0.05 nm$ is (wt. of one mole of proton $= 1.008 g$ )

$0.325 e V$

$5.205 e V$

$52.05 e V$

$3.25 e V$

Q. The accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.05nm is (wt. of one mole of proton =1.008g )

0.325eV

5.205eV

52.05eV

3.25eV

Q. The accelerating potential that must be imparted to a proton beam to give it an effective wavelength of $0.05 nm$ is (wt. of one mole of proton $= 1.008 g$ )

$0.325 e V$

$5.205 e V$

$52.05 e V$

$3.25 e V$