Question

An electron has a kinetic energy of 4.10 eV. Find its wavelength. 5.995 times 10^23 - 1 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully A photon has energy 4.10 eV. Find its wavelength 3.03 times 10^-3 Your response is off by a multiple of ten, 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 moving? 10 5 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

Answer 1

Use the DeBroglie relation

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK.E}}$

a] For the electron: K.E = 4.10 eV

so, $\lambda = \frac{6.626\times 10^{-34}}{\sqrt{2(9.1\times 10^{-31})(4.1\times 1.6\times 10^{-19})}} = 6.064\times 10^{-10}m$

b] For the proton: K.E = 4.10 eV

so, $\lambda = \frac{6.626\times 10^{-34}}{\sqrt{2(1.67\times 10^{-27})(4.1\times 1.6\times 10^{-19})}} = 1.415\times 10^{-11}m$

2]

Use Heisenberg's uncertainty relation:

$\Delta x \Delta p \geq \hbar$

=> $\Delta x [m\Delta v ] \geq \hbar$

=> $v\approx \Delta v = \frac{\hbar}{m\Delta x }$

=> $v\approx \frac{1.05456\times 10^{-34}}{(9.1\times 10^{-31})0.1\times 10^{-9} }=1.16\times 10^6 m/s$ or about 10⁶ m/s.