Question

in a photoelectric experiment the potential required to stop the ejection of electrons from cathode is 4v what is the value of maximum kinetic energy of emitted photoelectron ​

Answer 1

Answer:

Given

a

2

=13 and a

5

=25

We know that

a

n

=a+(n−1)d

∴a

2

=a+(2−1)d

⇒13=a+(2−1)d

⇒a+d=13 ...(i)

and a

5

=a+(5−1)d

⇒25=a+4d

⇒a+4d=25 ...(ii)

Now, subtracting (i) from (ii), we get

3d=12

⇒d=4

Now a+d=13 [from (i)]

⇒a=9

Hence, a

7

=a+6d=9+6(4)

⇒a

7

=33