Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
is the wavelength of photon of energy 35 Kev. h=6.625 × 10-34 J -s, c=3 × 108 m / s , 1 eV =1.6 × 10-19 J.
Q. is the wavelength of photon of energy
35
Ke
v
.
h
=
6.625
×
1
0
−
34
J
−
s
,
c
=
3
×
1
0
8
m
/
s
,
1
e
V
=
1.6
×
1
0
−
19
J
.
1244
182
Gujarat CET
Gujarat CET 2018
Report Error
A
35
×
1
0
−
12
m
B
35
A
˚
C
3.5
nm
D
3.5
A
˚
Solution:
λ
=
E
λ
h
c
=
355
×
1
0
3
×
1.6
×
1
0
−
19
6.625
×
1
0
−
34
×
3
×
1
0
8
m
=
0.35
×
1
0
−
10
m
=
3
×
1
0
−
12
m