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1 year ago

Please help help help mee

Physics

1 answer:

vampirchik [111]1 year ago

4 0

Answer:

.60 A

Explanation:

The RMS (Root Mean Square) voltage (the effective voltage) of the PEAK voltage is .7 * peak

.7 * 425 = 297.5 v

Now, V = IR re-arranged to V/R = I

297.5 v / 500 ohm = .595 A = ~ .6 A

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What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

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Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

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$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

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Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

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3 years ago

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Answer:

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KE = ½ (15 kg) (-20 m/s)²

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3 0

3 years ago