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Chapter 37: Q. 17 (page 1082)

Express in eV(or keVor MeVif more appropriate):

a. The kinetic energy of an electron moving with a speed of 5.0脳106m/s.

b. The potential energy of an electron and a proton 0.10nmapart.

c. The kinetic energy of a proton that has accelerated from rest through a potential difference of 5000V.

Short Answer

Expert verified

a. The kinetic energy of an electron isEk=71.17eV

b. The potential energy of an electron and a proton 0.10nmapart isU=-14.42eV.

c. The kinetic energy of a proton isEk=5keV.

Step by step solution

01

Part (a) step 1: Given information

We have given that the express in eV.

We need to find kinetic energy of an electron moving with a speed of5.0脳106m/s.

02

Part (a) step 2: Simplify

The expression for kinetic energy

Ek=12mv2Ek=12(9.11)(10-31)(5(106))2 Ek=1.13875(10-17)J

03

Part (b) step 1: Given information 

We have given that the express ineV

We need to find potential energy of an electron and a proton 0.10nmapart.

04

Part (b) step 2: Simplify

Using formula for potential energy :

U=q1q24蟺鈭�0rU=e(-e)4蟺鈭�0(0.1)(10-9)U=-2.307(10-18)J=-14.42eV

05

Part (c) step 1: Given information

We have given that the express in eV

We need to find the kinetic energy of a proton that has accelerated from rest through a potential difference of5000V.

06

Part (c) step 2: Simplify

The expression for kinetic energy when charge accelerating with a potential difference 鈭�V:

Ek=q鈭�VEk=e鈭�VEk=e(5000)Ek=5keV

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