Physics Questions - Need Help With a Couple (1 Viewer)

Life'sHard · Jul 19, 2021

1)

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2)

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3)

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4)

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5)

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notme123 · Jul 19, 2021

1)
$E=hf$
$E=\frac{hc}{ \lambda}$
$\therefore E=4.18*10^{-19} J$
B

2)
$f = 75*10^{6} Hz$
$\therefore \lambda=\frac{c}{f} = 4m$

$P -> R = 30m, P->Q -> R = 38m$
$\Delta{x} = 38-30 = 8m$
$\therefore \Delta{x} = 2(4) = 2\lambda$
B

3)
Rydeberg Equation
$\frac{1}{\lambda} = R(\frac{1}{n_{f}^2 }- \frac{1}{n_i^2})$ where $n_{f} =2$ and $n_i = 5$
$\therefore \frac{1}{\lambda} = R(\frac{1}{2^2} - \frac{1}{5^2 }) = 1.097 *10^7 (\frac{21}{100}) = 23037700 m^{-1}$
$\therefore \lambda = 434.08 *10^{-9} m$

D

4)
$F_{c}= F_{B}$
$\frac{mv^2}{r} = qvB\sin{90}$ because the proton is in circular motion when moving in a perpendicular mag feild.
$B= \frac{mv^2}{rqv} = \frac{mv}{rq}$
All these values are known: r and v are given in the question and q and m are on the formula sheet.
$\therefore B = 0.0159T$ which links with none of the answers please tell me if I've done something wrong.

5)
In the first diagram.
$I = I_0 \cos^{2}\theta$
$2 = 8 \cos^{2}\theta$
$\cos^{2}\theta = \frac{2}{8} = \frac{1}{4}$
$\cos\theta = \frac{1}{2}$
$\therefore \theta = 60$

Now in the second diagram:
$I = I_0 \cos^{2}(\theta + 90)$
$I = 8\cos^{2}(60 + 90) = 8\cos^{2}(150) = 8*\frac{3}{4} = 6 W/m^2$

C

Life'sHard · Jul 19, 2021

Geez tyvm

Physics Questions - Need Help With a Couple (1 Viewer)

Life'sHard

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notme123

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Life'sHard

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