menu search
brightness_auto
Feel free to answer or ask any questions.
If you are new here please see how to use or FAQ.
more_vert

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is :

(me = 9 × 10−31 kg, e = 1.6 × 10−19 C)

(Take $1\over4\pi\epsilon_o$ = 9 × 109Nm2C2)

(1) 1025 m/s2

(2) 1024 m/s2

(3) 1023 m/s2

(4) 1022 m/s2

thumb_up_off_alt 1 like thumb_down_off_alt 0 dislike

1 Answer

more_vert
 
done_all
Best answer

Answer (4): 1022 m/s2

Solution:

$F = {1\over 4\pi e_0} × {e^2\over r}$

= 9 × 10-9

Acceleration of electron and proton = $F\over Me$

=1022 m/s2

thumb_up_off_alt 1 like thumb_down_off_alt 0 dislike
Welcome to Jee Neet QnA, where you can ask questions and receive answers from other members of the community.

1.1k questions

771 answers

366 comments

77 users

...