The circumference of the second orbit of electron in hydrogen atom is 400 nm, the de-Broglie wavelength of election corresponding to the circumference of same orbitis 200 rm Number of waves made by an election ish
Answers
Answered by
0
Answer:
Number of waves 'n' =
Wavelength
Circumference
nλ=2πr
2λ=600
λ=300nm
Let stopping potential is V
0
.
eV
0
=
2
1
mv
2
...(i)
λ=
mv
h
v=
λm
h
...(ii)
From equations (i) and (ii),
eV
0
=
2
1
m(
λm
h
)
2
V
0
=
2mλ
2
e
h
2
=
2×(9.1×10
−31
)×(300×10
−9
)
2
×1.6×10
−19
(6.626×10
−34
)
2
=1.675×10
−5
V.
Hence the potential difference to which the electron has to be subjected so that electron stops is =1.675×10
−5
V.
Similar questions