Answer:
0.565 nm (3 s.f.)
Explanation:
De Broglie Wavelength Formula
[tex]\lambda=\dfrac{h}{p}=\dfrac{h}{mv}[/tex]
where:
- λ = the de Broglie wavelength (m)
- h = Planck's constant (J s)
- p = momentum of the particle (kg m/s)
- m = mass of the particle (kg)
- v = speed of the particle (m/s)
Planck's Constant
A constant relating the energy of a photon to its frequency:
[tex]\sf h = 6.6261 \times 10^{-34} J\:s[/tex]
Given:
- v = 7.00 × 10² m/s
- m = 1.675 × 10⁻²⁷ kg
Substitute the given values into the formula (along with Planck's Constant):
[tex]\implies \lambda=\sf \dfrac{6.6261 \times 10^{-34}}{(1.675 \times 10^{-27})(7.00 \times 10^2)}[/tex]
[tex]\implies \lambda=\sf \dfrac{6.6261 \times 10^{-34}}{1.675 \times 7.00\times 10^{-27}\times 10^2}[/tex]
[tex]\implies \lambda=\sf \dfrac{6.6261 \times 10^{-34}}{1.675 \times 7.00\times 10^{-27+2}}[/tex]
[tex]\implies \lambda=\sf \dfrac{6.6261 \times 10^{-34}}{11.725\times 10^{-25}}[/tex]
[tex]\implies \lambda=\sf \dfrac{6.6261}{11.725}\times \dfrac{10^{-34}}{10^{-25}}[/tex]
[tex]\implies \lambda=\sf 0.5651257...\times \dfrac{10^{-34}}{10^{-25}}[/tex]
[tex]\implies \lambda=\sf 0.5651257...\times 10^{-34-(-25)}[/tex]
[tex]\implies \lambda=\sf 0.5651257...\times 10^{-9}[/tex]
[tex]\implies \lambda=\sf 5.651257...\times 10^{-10}\:\:m[/tex]
To convert meters (m) to nanometers (nm), multiply by 10⁹:
[tex]\implies \lambda=\sf (5.651257...\times 10^{-10} \times 10^9)\:\:nm[/tex]
[tex]\implies \lambda=\sf (5.651257...\times 10^{-10+9})\:\:nm[/tex]
[tex]\implies \lambda=\sf (5.651257...\times 10^{-1})\:\:nm[/tex]
[tex]\implies \lambda=\sf 0.5651257...\:\:nm[/tex]
[tex]\implies \lambda=\sf 0.565\:\:nm\:\:(3\:s.f.)[/tex]
Exponent rules used
[tex]\textsf{Product Rule}: \quad \sf a^b \cdot a^c=a^{b+c}[/tex]
[tex]\textsf{Quotient Rule}: \quad \sf \dfrac{a^b}{a^c}=a^{b-c}[/tex]
