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Structure Of Atom

Question
CBSEENCH11005260
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A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

Solution

Power of the bulb = 100 watt
                          = 100 Js-1
Energy of one photon E = hv = fraction numerator h c over denominator straight lambda end fraction
or space straight E space equals space fraction numerator 6.626 space cross times 10 to the power of negative 34 end exponent Js cross times 3 cross times 10 to the power of 8 ms to the power of negative 1 end exponent over denominator 400 cross times 10 to the power of negative 9 end exponent straight m end fraction space space space space equals space 4.969 space cross times 10 to the power of negative 19 end exponent straight J
therefore space space Number space of space photons space emitted space space space space equals fraction numerator Power over denominator Energy space of space photon end fraction space equals space fraction numerator 100 space Js to the power of negative 1 end exponent over denominator 4.969 space cross times space 10 to the power of negative 19 end exponent straight J end fraction space space space space equals space 2.012 space cross times space 10 to the power of 20 straight s to the power of negative 1 end exponent

Some More Questions From Structure of Atom Chapter

Name the fundamental particles which constitute an atom.

What is an Electron?

What is a proton?

Who discovered neutron?

What is a neutron?

What is the name of the remaining part of atom except outer orbit?

Name the particles which determine the mass of an element.

What are α-particles?

Why were neutrons discovered very late?

What are the fundamental particles present in a neutral atom having atomic number greater than 1?

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