Singly charged sodium ions are accelerated through a potential difference of 300 V. (a) What is the momentum acquired by such an ion? (b) What is its de Broglie wavelength?

Short Answer

Expert verified

(a) The momentum is 1.91×1021kgms.

(b) De Broglie wavelength is 3.46×1013 m.

Step by step solution

01

The given data:

Singly charged sodium ions are accelerated through a potential difference of300V.

02

Definition of de Broglie wavelength:

The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength.

The de Broglie wavelength is defined by using following formula.

λ=hp

Here, his Planck’s constant and pis the momentum.

Kinetic energy of ion is,

K=qV

Here, q is the charge having a value 1.6×1019 Cand Vis the potential having a value 300 V.

Write the equation for the momentum as,

p=2mK

Here, m is mass of ion and K is Kinetic energy.

03

(a) Determining the momentum of ion:

The kinetic energy of the ion is,

K=qV=(1.6×1019 C)(300 V)=4.8×1017 J

As the mass of single sodium ion is 3.819×1026 kg.

Thus, momentum of a sodium ion is,

role="math" localid="1663135253292" p=2mK=2(3.819×1026 kg)(4.8×1017 J)=1.91×1021kgms

Hence, the momentum of ion is 1.91×1021kgms.

04

(b) Determining the de Broglie wavelength:

The de Broglie wavelength is,

λ=hp=6.63×1034 Js1.91×1021  kgms=3.46×1013 m

Hence, de Broglie wavelength is 3.46×1013 m.

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