MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Formation of Ions in Metals

28.whar would...

Question

28.whar would be the distance of closest approach when an alpha particle of kinetic energy 5.5 MeV is bombarded to a gold nucleus.

Open in App

Solution

Suggest Corrections

thumbs-up

19

Similar questions

Q. The maximum K.E found in alpha particle of natural origin is

7.7

MeV. Find the closest distance of approach for the gold nucleus

(Z = 79

)

[Note : At closest distance of approach, kinetic energy is converted into potential energy]

Q. The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy

K

r_{0}

. The distance of the closest approach when the

α

particle is fired at the same nucleus with kinetic energy

2 K

Q. In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of

Z = 75

α

5 M e V

energy impinges on it before it comes momentarily to rest and reverse its direction.
How will the distance of closest approach be affected when the kinetic energy of the

α

-particle is doubled?

α - p a r t i c l e

5 M e V

is scattered through

180^{\circ}

by gold nucleus. The distance of closest approach is of the order of

α

-particles with initial kinetic energy

4.8

mev are shot at gold atoms

(z = 79)

. The potential of

α

-particle at the closest distance to the gold nucleus is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Speed Math

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Formation of Ions in Metals

Standard VIII Chemistry

Join BYJU'S Learning Program

CrossIcon