# 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

P perimeter

As we learnt in

Energy of a photon -

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

- wherein

$\fn_jvn h= Plank's\: constant$

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

$\fn_jvn \lambda \rightarrow wave \: length$

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

$\fn_jvn h= Plank's\: constant$

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

$\fn_jvn \lambda \rightarrow wave \: length$

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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