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Standard XII
Physics
NCERT
Question
Value of E(in eV) is
32.2
24.6
20.8
23.8
A
32.2
B
24.6
C
23.8
D
20.8
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Solution
Verified by Toppr
Ionization Energy of atom of atomic number
Z
in
n
th orbit is
13.6
Z
2
/
n
2
. So work function for metal A
W
A
=
13.6
/
2
2
=
3.4
e
V
and,
work function for metal B
W
B
=
13.6
×
2
2
/
2
2
=
13.6
e
V
Now,
E
−
W
A
=
K
E
A
and,
E
−
W
B
=
K
E
B
given that,
K
E
A
=
2
K
E
B
so,
E
−
W
A
=
2
E
−
2
W
B
⇒
E
=
2
W
B
−
W
A
=
27.2
−
3.4
=
23.8
e
V
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