If a source of power 4 kW produces 1020 photons/second, the radiation belongs to a part of the spectrum called

Option 1)

\gamma -rays\;

Option 2)

\; X-rays

Option 3)

ultraviolet rays

Option 4)

microwaves

Answers (2)
P perimeter

As we learnt in

Energy of a photon -

\fn_jvn E= h\nu = \frac{hc}{\lambda }

- wherein

h= Plank's\: constant

\boldsymbol{\nu= frequency\: of \: radiation }

\lambda \rightarrow wave \: length

 

 If wavelength of photon is {\lambda} & n is the number of photon emitted per second then

P=n\frac{hc}{\lambda}

4\times10^{3}=\frac{10^{20}\times 6.62\times{10^{-34}\times{3}\times{10^{8}}}}{\lambda}

{\lambda}=\frac{19.8\times{10^{-26}\times{10^{20}}}}{4\times{10^{3}}}=4.96\times{10^{-9}}

{\lambda}=49.6\AA

This wavelength represent X-rays hence correct answer is option 2.


Option 1)

\gamma -rays\;

This is an incorrect option.

Option 2)

\; X-rays

This is the correct option.

Option 3)

ultraviolet rays

This is an incorrect option.

Option 4)

microwaves

This is an incorrect option.

P perimeter

As we learnt in

Energy of a photon -

\fn_jvn E= h\nu = \frac{hc}{\lambda }

- wherein

h= Plank's\: constant

\boldsymbol{\nu= frequency\: of \: radiation }

\lambda \rightarrow wave \: length

 

 If wavelength of photon is {\lambda} & n is the number of photon emitted per second then

P=n\frac{hc}{\lambda}

4\times10^{3}=\frac{10^{20}\times 6.62\times{10^{-34}\times{3}\times{10^{8}}}}{\lambda}

{\lambda}=\frac{19.8\times{10^{-26}\times{10^{20}}}}{4\times{10^{3}}}=4.96\times{10^{-9}}

{\lambda}=49.6\AA

This wavelength represent X-rays hence correct answer is 2


Option 1)

\gamma -rays\;

This is an incorrect option.

Option 2)

\; X-rays

This is the correct option.

Option 3)

ultraviolet rays

This is an incorrect option.

Option 4)

microwaves

This is an incorrect option.

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