Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
The energy of one mole of photons of radiation of wavelength 300 nm is (Given: h =6.63 × 10-34 Js , N A =6.02 × 1023 mol -1, c =3 × 108 ms -1 )
Q. The energy of one mole of photons of radiation of wavelength
300
nm
is
(Given :
h
=
6.63
×
1
0
−
34
J
s
,
N
A
=
6.02
×
1
0
23
m
o
l
−
1
,
c
=
3
×
1
0
8
m
s
−
1
)
1058
166
JEE Main
JEE Main 2022
Structure of Atom
Report Error
A
235
k
J
m
o
l
−
1
5%
B
325
k
J
m
o
l
−
1
6%
C
399
k
J
m
o
l
−
1
84%
D
435
k
J
m
o
l
−
1
5%
Solution:
Energy of one mole of photons
=
λ
h
c
×
N
A
=
300
×
1
0
−
9
6.63
×
1
0
−
34
×
3
×
1
0
8
×
6.02
×
1
0
23
=
399.13
×
1
0
3
Joule / mole
=
399
kJ / mole