4) Calculate the kinetic energy of moving electron
which has a wavelength of 4.8 pm.
(mass of electron = 9.11 x 10-31 kg,
h = 6.63 x 10-34 J s).
a) 10^-10J
b) 10^-12J
c) 10^-14J
d) 10^-16J
Answers
Answer:
Given :
Wavelength of electron = 4.8 pm
To find :
The Kinetic energy of the moving electron
Solution :
The relation between picometres and metres is given by ,
\boxed {\rm{1pm = {10}^{ - 12} \: m}}
1pm=10
−12
m
: \implies \rm \: 4.8 \: pm = 4.8 \times {10}^{ - 12} \: m:⟹4.8pm=4.8×10
−12
m
The relation between planck's constant , wavelength , kinetic energy and mass is given by ,
\dag \: \boxed {\rm{ \lambda = \frac{h}{ \sqrt{2mKE} } }}†
λ=
2mKE
h
We have ,
mass of electron = 9.1 × 10⁻³¹ kg
KE = 4.8 × 10⁻¹² m
h = 6.63 × 10⁻³⁴ J.S
By substituting the values ,
\begin{gathered} : \implies \rm \: 4.8 \times {10}^{ - 12} \: m = \frac{6.63 \times {10}^{ - 34} \: J.s}{ \sqrt{2 \times 9.1 \times {10}^{ - 31} \: kg \times KE} } \\ \\ : \implies \rm \: 4.8 \times {10}^{ - 12} \: m = \frac{6.63 \times {10}^{ - 34} \: kgm {s}^{ - 2}.s }{ \sqrt{18.2 \times {10}^{ - 31} kg \times KE} } \\ \\ : \implies \rm \: 4.8 \times {10}^{ - 12} \: m \times \sqrt{18.2 \times {10}^{ - 31} \: kg \times KE } = 6.63 \times {10}^{ - 34} \: kg.m {s}^{ - 1} \\ \\ : \implies \rm \: 4.8 \times {10}^{ - 12} \: m \times 1.34 \times {10}^{ - 15} \: kg {}^{1 \div 2} \times \sqrt{KE} = 6.63 \times {10}^{ - 34} \: kg.m {s}^{ - 1} \\ \\ : \implies \rm \: 6.432 \times {10}^{ - 27} \: kg {}^{1 \div 2} .m \times \sqrt{KE} = 6.63 \times {10}^{ - 34} \: kgm {s}^{ - 1} \\ \\ : \implies \rm \: \sqrt{KE} = \frac{6.63 \times {10}^{ - 34} \: kgm {s}^{ - 1} }{6.432 \times {10}^{ - 27} \: kg {}^{1 \div 2}.m } \end{gathered}
:⟹4.8×10
−12
m=
2×9.1×10
−31
kg×KE
6.63×10
−34
J.s
:⟹4.8×10
−12
m=
18.2×10
−31
kg×KE
6.63×10
−34
kgms
−2
.s
:⟹4.8×10
−12
m×
18.2×10
−31
kg×KE
=6.63×10
−34
kg.ms
−1
:⟹4.8×10
−12
m×1.34×10
−15
kg
1÷2
×
KE
=6.63×10
−34
kg.ms
−1
:⟹6.432×10
−27
kg
1÷2
.m×
KE
=6.63×10
−34
kgms
−1
:⟹
KE
=
6.432×10
−27
kg
1÷2
.m
6.63×10
−34
kgms
−1
\begin{gathered} : \implies \rm \: KE= \bigg(\frac{6.63 \times {10}^{ - 34} \: kg.m {s}^{ - 1} }{6.432 \times {10}^{ - 27} \: kg {}^{1 \div 2} m} \bigg)^{2} \\ \\ : \implies \rm \: KE = \frac{(6.63 \times {10}^{ - 34} \: kgm {s}^{ - 1} ) {}^{2} }{(6.432 \times {10}^{ - 27} \: kg {}^{1 \div 2} m) {}^{2} } \\ \\ : \implies \rm \: KE= \frac{4.39 \times 10 {}^{ - 67} \: kg {}^{2} {m}^{2} {s}^{ - 2} }{4.13 \times {10}^{ - 53} \: (kg {}^{1 \div 2} ) {}^{2}m {}^{2} } \\ \\ : \implies \rm \: KE = \frac{4.39 \times 10 {}^{ - 14} \: kg {}^{2}m {}^{2} {s}^{ - 2} }{4.13 \: kgm {}^{2} } \end{gathered}
:⟹KE=(
6.432×10
−27
kg
1÷2
m
6.63×10
−34
kg.ms
−1
)
2
:⟹KE=
(6.432×10
−27
kg
1÷2
m)
2
(6.63×10
−34
kgms
−1
)
2
:⟹KE=
4.13×10
−53
(kg
1÷2
)
2
m
2
4.39×10
−67
kg
2
m
2
s
−2
:⟹KE=
4.13kgm
2
4.39×10
−14
kg
2
m
2
s
−2
\begin{gathered} : \implies \rm \: KE= \frac{4.39 \times {10}^{ - 14} \: kg.m {s}^{ - 2} }{4.13} \\ \\ : \implies \rm \: KE = 1.06 \times {10}^{ - 14} \: kg.m {s}^{ - 2} \\ \\ : \implies \rm \: KE = 1.06 \times {10}^{ - 14} \: J\end{gathered}
:⟹KE=
4.13
4.39×10
−14
kg.ms
−2
:⟹KE=1.06×10
−14
kg.ms
−2
:⟹KE=1.06×10
−14
J
Hence , The kinetic energy of the moving electron is 1.06 × 10⁻¹⁴ J