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OLga [1]
2 years ago
14

PLEASE HELP ME WITH THIS ONE QUESTION

Physics
1 answer:
melamori03 [73]2 years ago
3 0

Explanation:

First, we convert the energy from eV to Joules:

2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)

= 4.64×10^{-19}\:\text{eV}

We know from definition that

E=h\nu = \dfrac{hc}{\lambda}

so the wavelength of the photon is

\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}

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When they are connected in series

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        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

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