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2 years ago

What is the energy of a photon of violet light with a frequency of 7.57x10^14 s^-1?

Chemistry

1 answer:

Eddi Din [679]2 years ago

6 0

Answer:

ΔE = 5.02 x 10⁻¹⁹ j

Explanation:

ΔE (photon) = h·f = (6.63 x 10⁻³⁴ j·s)(7.57 x 10¹⁴ s⁻¹) = 5.02 x 10⁻¹⁹ j

h = Planck's Constant = 6.63 x 10⁻³⁴ j·s

f = frequency (given) = 7.57 x 10¹⁴ s⁻¹

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3 years ago

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1 year ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

Answer:

For a: The edge length of the crystal is 533.5 pm

For b: The atomic radius of potassium is 231.01 pm

Explanation:

For a:

To calculate the lattice parameter or edge length of the crystal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $0.855g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal = 39.09 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = ?

Putting values in above equation, we get:

$0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm$

Conversion factor: $1cm=10^{10}pm$

Hence, the edge length of the crystal is 533.5 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm$

Hence, the atomic radius of potassium is 231.01 pm

4 0

3 years ago