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3 years ago

How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 39 j of work?

1 answer:

3 0

The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
$W=q \Delta V$
where q is the charge and $\Delta V$ is the potential difference.

In our problem, the work done is W=39 J while the potential difference of the battery is $\Delta V = 9.0 V$ , so we can find the charge transferred by the battery:
$q= \frac{W}{\Delta V}= \frac{39 J}{9.0 V}=4.33 C$

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notka56 [123]

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

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svp [43]

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