Answer:
0.565 nm (3 s.f.)
Step-by-step explanation:
De Broglie Wavelength Formula
![\lambda=(h)/(p)=(h)/(mv)](https://img.qammunity.org/2023/formulas/chemistry/high-school/9nt7d5h477jp1ez1ugqmg01etgno1ooa6m.png)
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:
![\sf h = 6.6261 * 10^(-34) J\:s](https://img.qammunity.org/2023/formulas/chemistry/high-school/5ixghqj8hqzhbgy41vgs5ikbgcl3nay97p.png)
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):
![\implies \lambda=\sf (6.6261 * 10^(-34))/((1.675 * 10^(-27))(7.00 * 10^2))](https://img.qammunity.org/2023/formulas/chemistry/high-school/1mlbix0qv810gicru27ytu9yuphg2rfmwb.png)
![\implies \lambda=\sf (6.6261 * 10^(-34))/(1.675 * 7.00* 10^(-27)* 10^2)](https://img.qammunity.org/2023/formulas/chemistry/high-school/zbl3es9m2n0qokra9fb6rq2xs1dgvf7t8a.png)
![\implies \lambda=\sf (6.6261 * 10^(-34))/(1.675 * 7.00* 10^(-27+2))](https://img.qammunity.org/2023/formulas/chemistry/high-school/dqfteu4bxpgrrclswvsp60toifvjq5rrfi.png)
![\implies \lambda=\sf (6.6261 * 10^(-34))/(11.725* 10^(-25))](https://img.qammunity.org/2023/formulas/chemistry/high-school/rhk9ijh7l8lsrtixzp1bkokmcltrlch9v8.png)
![\implies \lambda=\sf (6.6261)/(11.725)* (10^(-34))/(10^(-25))](https://img.qammunity.org/2023/formulas/chemistry/high-school/p6e3mvdwhrec5oy9ctou0zjj22nkb938r0.png)
![\implies \lambda=\sf 0.5651257...* (10^(-34))/(10^(-25))](https://img.qammunity.org/2023/formulas/chemistry/high-school/i7ldm7w3e8twdonpkdvaal9r02t8pea8cn.png)
![\implies \lambda=\sf 0.5651257...* 10^(-34-(-25))](https://img.qammunity.org/2023/formulas/chemistry/high-school/76jom6v0qop491b596r9piuew81ivx4suk.png)
![\implies \lambda=\sf 0.5651257...* 10^(-9)](https://img.qammunity.org/2023/formulas/chemistry/high-school/v7u44d5jgc0wpet7n8xd10kq04rcnfiuru.png)
![\implies \lambda=\sf 5.651257...* 10^(-10)\:\:m](https://img.qammunity.org/2023/formulas/chemistry/high-school/zmlfag53qp81mt4gv8oayllb7okwq7s2w5.png)
To convert meters (m) to nanometers (nm), multiply by 10⁹:
![\implies \lambda=\sf (5.651257...* 10^(-10) * 10^9)\:\:nm](https://img.qammunity.org/2023/formulas/chemistry/high-school/2cmo3mhv2b8emaoeu4956a3772ol2gokf1.png)
![\implies \lambda=\sf (5.651257...* 10^(-10+9))\:\:nm](https://img.qammunity.org/2023/formulas/chemistry/high-school/jwn75346t6b0x2tenv1p99h811yyrmsen9.png)
![\implies \lambda=\sf (5.651257...* 10^(-1))\:\:nm](https://img.qammunity.org/2023/formulas/chemistry/high-school/7o5yyx58n5hv2viy88ch6b39kqylgfj7yr.png)
![\implies \lambda=\sf 0.5651257...\:\:nm](https://img.qammunity.org/2023/formulas/chemistry/high-school/thyiijjzi9799qq2tkmroz00ddx3wj6xcn.png)
![\implies \lambda=\sf 0.565\:\:nm\:\:(3\:s.f.)](https://img.qammunity.org/2023/formulas/chemistry/high-school/2bv5e5k547c2jc1dzfauw8gkzajmycmm29.png)
Exponent rules used
![\textsf{Product Rule}: \quad \sf a^b \cdot a^c=a^(b+c)](https://img.qammunity.org/2023/formulas/chemistry/high-school/np37i7whd4ix1v2ryctmych5a3g2hp9rzy.png)
![\textsf{Quotient Rule}: \quad \sf (a^b)/(a^c)=a^(b-c)](https://img.qammunity.org/2023/formulas/chemistry/high-school/t1iotvzghfr8cq9tsozllk3wrqhdttgigp.png)