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aleksley [76]
2 years ago
8

Since many wavelengths of light are measured in nanometers, it's useful to know that planck's constant (h) multiplied with the s

peed of
light (c) is equal to 1240 eVnm.
This is useful, as E = hf, or otherwise written as E= .
With this information in mind, solve the following:
A photon has 1.6 eV of energy. What is the wavelength of this photon?
O a
410 nm
Ob
775 nm
Ос
1984 nm
Od
306 nm
Physics
1 answer:
MrRa [10]2 years ago
4 0

Answer:

E = h f      and since f = c / λ     we have E = h c / λ

Thus  λ = h c / E

Also, 1 ev = 1.6E-19 C

λ = 6.63E-34 * 3.0E8 / (1.6 * 1.6E-19)

λ = 6.63 * 3.0 / (1.6 * 1.6) * E-7

λ = 7.77E-7 = 777 nanometers

(B) is correct

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7 0
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What is the density of a cube that has a mass of 3.75 g and a volume of 3 mL?
valina [46]

Answer:

\displaystyle \rho=1.25\ g/ml

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

\displaystyle \rho=\frac{m}{V}

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\displaystyle \rho=\frac{3.75\ g}{3\ ml}

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6 0
2 years ago
The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E =  K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

 E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0
3 years ago
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