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Explain the method to decrease friction?
2 answers:
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Answer:
9.430 * 10¹⁷ protons per second whill shine on the book from a 62 W bulb
Explanation:
To answer this question, first let's calculate the energy of a single photon with a wavelength (λ) of 504 nm:
E = hc/λ
Where h is Planck's constant (6.626*10⁻³⁴ J·s) and c is the speed of light (3*10⁸ m/s).
E = 6.626*10⁻³⁴ J·s * 3*10⁸ m/s ÷ (504*10⁻⁹m) = 3.944 * 10⁻¹⁹ J.
So now we can make the equivalency for this problem, that
<u>1 proton = 3.944 * 10⁻¹⁹ J</u>
Now we convert watts from J/s to proton/s:
1 = 1 W
Solving the problem, a 62 W bulb converts 5% of its output into light, so:
- 62 * 5/100 = 3.1 W
3.1 watts are equal to [ 2.535*10¹⁸ proton/s * 3.1 ] = 7.858 * 10¹⁸ proton/s
Of those protons per second, 12% will shine on the chemistry textbook, thus:
7.858 * 10¹⁸ proton/s * 12/100 = 9.430 * 10¹⁷ protons/s