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

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a