Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed of 8.83 × 106 m/s and (b) a kinetic energy of 7.81 × 10−15 J. Enter your answers in scientific notation.

a) Wavelength of electron = 8.21 × 10⁻¹¹ m = 821 pm

Wavelength of proton = 4.50 × 10⁻¹³ m = 0.45 pm

The electron has a higher wavelength than the electron.

b) Wavelength of electron = 4.61 × 10⁻¹⁰ m = 461 pm

Wavelength of proton = 1.30 × 10⁻¹³ m = 0.130 pm

The electron once again has a higher wavelength.

Note 1 pm = 10⁻¹² m

Explanation:

a) The relationship between wavelength, mass of a particle and its speed is given in the De Broglie's equation

λ = h/mv

For the two particles

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

v = 8.83 × 10⁶ m/s

For an electron

m = 9.11 × 10⁻³¹ kg

Wavelength = (6.63 × 10⁻³⁴)/(9.11 × 10⁻³¹ × 8.83 × 10⁶) = 8.21 × 10⁻¹¹ m = 821 pm

For a proton

m = 1.67 × 10⁻²⁷ kg

Wavelength = (6.63 × 10⁻³⁴)/(1.67 × 10⁻²⁷ × 8.83 × 10⁶) = 4.50 × 10⁻¹³ m = 0.45 pm

b) The velocity needs to be first obtained from the kinetic energy relation. Before using the De Broglie's equation.

K.E = mv²/2

v = √(2K.E/m)

For the electron,

m = 9.11 × 10⁻³¹ kg

v = √[(2 × 7.81 × 10⁻¹⁵)/(9.11 × 10⁻³¹)]

v = 1.31 × 10⁸ m/s

Wavelength = h/mv

Wavelength = (6.63 × 10⁻³⁴)/(9.11 × 10⁻³¹ × 1.31 × 10⁸) = 4.61 × 10⁻¹⁰ m = 461 pm

For the proton,

m = 1.67 × 10⁻²⁷ kg

v = √[(2 × 7.81 × 10⁻¹⁵)/(1.67 × 10⁻²⁷)]

v = 3.06 × 10⁶ m/s

Wavelength = h/mv

Wavelength = (6.63 × 10⁻³⁴)/(1.67 × 10⁻²⁷ × 3.06 × 10⁶) = 1.30 × 10⁻¹³ m = 0.130 pm

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

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