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enyata [817]
3 years ago
13

A photon has an energy of 5.53 à 10¯17 j. what is its frequency in s¯1?

Physics
1 answer:
Butoxors [25]3 years ago
5 0
<span>energy is directly proportional fo frequency 
as in blue photons are more energetic than red photons 
so 
E = h * f 
look up h (Planck's constant) 
plug and play!</span><span>
</span>
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A box with a mass of 2 kg accelerates in a straight line
mihalych1998 [28]

Answer: 16N

Explanation:

Given that:

mass of box M= 2 kg

Initial speed V1 = 4 m/s

Final speed V2 = 8 m/s

Time taken T= 0.5 s

Average strength of this force F = ?

Now, recall that Force is the rate of change of momentum per unit time

i.e Force = momentum / time

Hence, F = M x (V2 - V1)/T

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7 0
3 years ago
What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
PolarNik [594]

Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m

So,

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

Also, \Delta E=h\times \nu

So,

h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}

\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

8 0
3 years ago
1 7.1.2 Quiz: Electrostatics
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Answer:

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The electric force between the two particles are calculated using the formular:

F = kQ₁Q₂ / d²

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k= Coulomb's law constant.

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d= distance.

the equation above is called Coulomb's law.

It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.

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