KVPY SA Exam 2018 Question Paper

Question : 22 of 80

Marks: +1, -0

One can define an alpha–Volt (α V) to be the energy acquired by an α particle when it is accelerated by a potential of 1 Volt. For this problem you may take a proton to 2000 times heavier than an electron. Then

1 α V = 1 eV/4000
1 α V = 2 eV
1 α V = 8000 eV
1 α V = 1 eV

Validate

Solution:

q = +2e

1αV = + 2 ev.

Go to Question:

Prev Question

Next Question