If the wavelength of an electron is 4.82 times 10^-7 m, how fast is it moving? k

Question

If the wavelength of an electron is 4.82 times 10^-7 m, how fast is it moving? km/s (b) If the electron has a speed equal to 9.40 times 10^6 m/s, what is its wavelength? m (a) An electron has a kinetic energy of 1.94 eV. Find its wavelength, nm (b) A photon has energy 1.94 eV. Find its wavelength, nm In the ground state of hydrogen, the uncertainty in the position of the electron is roughly 0.10 nm. If the speed of the electron is approximately the same as the uncertainty in its speed, about how fast is it moving? 10 m/s

Explanation / Answer

8. a)given data
wavelength = 4.82*10^-7 m
speed = ?
The deBroglie wavelength is:
lambda = h/p = h / mv

Solve for the speed:
v = h / (lambda m)
v = 6.62*10^-34 /(4.82*10^-7*9.1*10^-31)=1509m/s = 1.51km/s
b)v= 9.40*10^6 m/s
wavelength=?
The deBroglie wavelength is:
lambda = h/p = h / mv
= 6.62*10^-34/(9.1*10^-31*9.40*10^6)
=7.74*10^-11m

9. a)
Given data
KE = 1.94 eV = 1.94*1.602*10^-19 J
wavelength =?
lambda = h/sqrt(2*m*KE)
lambda = 6.62*10^-34/sqrt(2*9.1*10^-31*1.94*1.602*10^-19)=8.8*10^-10 m
=0.88 nm
b)energy = 1.94 eV
lambda=?
lambda = 6.62*10^-34/sqrt(2*9.1*10^-31*1.94*1.67*10^-27)=8.862*10^-6 m
=8862nm

10.Given data
uncertainity in position =0.10nm=0.10*10^-9 m
The mass of an electron = 9.1* 10^-31 kg
Planck's constant =6.62 * 10^-34 Js.
Heisenberg Uncertainty Principle states that
xp h/4
or
xmv h/4
since p=mv and m is constant
v (6.62*10^-34)/(4*3.14*0.10*10^-9*9.1*10^-34)
v 5.8*10^8 m/s

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