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Question:-A Photon Particle Has Wavelength 4950 A.
What Is The Power Of the Photon In Volt.[h=6.62×10^-27 erg sec]
{In Photo It's In Bengali language}
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hello dear brother...!!
according to the given question , wavelength λ = 4950 A° and it is also given
h = 6.62 x 10∧ -34 J
now , to calculate power of the photon , the formula that can be used is
⇒ E = h c / λ
⇒ E = { 6.62 x 10∧ -34 x 3 x 10∧ 8 } / 4950 x 10∧ -10
⇒ E = { 19 . 86 x 10 ∧ -26 } / 4950 x 10 ∧ -10
⇒ E = 4.012 x 10∧ -19 J
now to convert this into volts ,
4.02 x 10∧ -19 J / 1.6 x 10∧ -19 = 2.512 eV
therefore the power of the photon is 2.512 eV .
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hope it is correct and useful to you...!!
according to the given question , wavelength λ = 4950 A° and it is also given
h = 6.62 x 10∧ -34 J
now , to calculate power of the photon , the formula that can be used is
⇒ E = h c / λ
⇒ E = { 6.62 x 10∧ -34 x 3 x 10∧ 8 } / 4950 x 10∧ -10
⇒ E = { 19 . 86 x 10 ∧ -26 } / 4950 x 10 ∧ -10
⇒ E = 4.012 x 10∧ -19 J
now to convert this into volts ,
4.02 x 10∧ -19 J / 1.6 x 10∧ -19 = 2.512 eV
therefore the power of the photon is 2.512 eV .
-------------------------------------------
hope it is correct and useful to you...!!
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