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What is the approximate wavelength of : a) an electron moving with energy 10 eV?

b) a photon with an elegy of 10 eV?
c) a neutron with an energy of 0.1 eV?
(Please explain)

User Heyarne
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1 Answer

2 votes

Answer:

(A)
\lambda =12.375* 10^(-8)m

(B)
\lambda =12.375* 10^(-8)m

(C)
\lambda =12.375* 10^(-6)m

Step-by-step explanation:

(a) We have given energy of the electron = 10 eV

We know that
1eV=1.6* 10^(-19)J

So 10 eV =
=10* 1.6* 10^(-19)J=1.6* 10^(-18)j

Speed of light
c=3* 10^8m/sec

Plank's constant
h=6.6* 10^(-34)J-s

Energy of electron is given by
E=h\\u =(hc)/(\lambda )


1.6* 10^(-18)=(6.6* 10^(-34)* 3* 10^8)/(\lambda )


\lambda =12.375* 10^(-8)m

(b) Energy of photon is also given as E =10 eV

We know that
1eV=1.6* 10^(-19)J

So 10 eV =
=10* 1.6* 10^(-19)J=1.6* 10^(-18)j

Energy of photon is given by
E=h\\u =(hc)/(\lambda )

Speed of light
c=3* 10^8m/sec

Plank's constant
h=6.6* 10^(-34)J-s


1.6* 10^(-18)=(6.6* 10^(-34)* 3* 10^8)/(\lambda )


\lambda =12.375* 10^(-8)m

(c) Energy of neutron is given as E= 0.1 eV

We know that
1eV=1.6* 10^(-19)J

So 0.1 eV =
=10* 1.6* 10^(-19)J=1.6* 10^(-20)j

Speed of light
c=3* 10^8m/sec

Plank's constant
h=6.6* 10^(-34)J-s


1.6* 10^(-20)=(6.6* 10^(-34)* 3* 10^8)/(\lambda )


\lambda =12.375* 10^(-6)m

User AvaTaylor
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