Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.

1 answer:

8 0

The energy of a photon is given by
$E=hf$
where
$h=6.6 \cdot 10^{-34} Js$ is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is
$f=5.49 \cdot 10^{14}Hz$
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
$E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J$

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An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one

aalyn [17]

Answer:

20.78 m/s that we can approximate to option d (21 m/s)

Explanation:

The solution involves a lot of algebra and to be familiar with different convenient formulas for launching an object vertically under the action of gravity.

First you need to recall (or derive) the formula for the maximum height reached by an object with launches with initial velocity $v_0$ :

Maximum height = $\frac{{v_0}^2}{2g}$

Therefore one fourth of such height would be: $\frac{{v_0}^2}{8g}$

Second, find what would be the time needed to reach that height by solving for the time in the equation for the vertical position:

$y(t)=v_0*t-\frac{g}{2} t^2 \\\frac{{v_0}^2}{8g} = v_0*t-\frac{g}{2} t^2\\\frac{g}{2} t^2-v_0*t+\frac{{v_0}^2}{8g}=0$

And now, solve for t in the last equation using the quadratic formula to find the time needed for the object to reach that height (one fourth of the max height):

$t=\frac{v_0+/-\sqrt{{v_o}^2-4*\frac{g}{2}*\frac{{v_o}^2}{8g} } }{g} = \frac{v_0+/-\sqrt{{v_o}^2-\frac{{v_o}^2}{4} } }{g} =\frac{v_0+/-\sqrt{\frac{3{v_o}^2}{4} } }{g} = \\=\frac{v_0+/-v_0\sqrt{\frac{3}{4} } }{g} =\frac{v_0+/-v_0\frac{\sqrt{3}}{2} } {g}$

Next, use this expression for t in the equation for the velocity at any time t in the object's trajectory that comes from the definition of acceleration;

$v(t)=v_0-g*t$