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11.Dual Nature of Radiation and matter
easy
જો ફોટોનની ઊર્જા $KeV $ એકમમાં અને તરંગલંબાઈ $Å$ એકમમાં દર્શાવવામાં આવે, તો ફોટોનની ઊર્જા ............ દ્ધારા શોધી શકાય.
A$E = 12.4 \,hf$
B$E = 12.4 h/\lambda$
C$E = 12.4/\lambda$
D$E = hf$
Solution
ફોટોનની ઊર્જા ${\text{E}}\,\, = \,\,\frac{{{\text{hc}}}}{\lambda }$(જૂલ એકમમાં )
$ = \,\,\frac{{hc}}{{e\lambda }}$ ($eV\,$ એકમમાં)
${E_{(eV)}}\, = \,\,\,\frac{{6.6\,\, \times \,\,{{10}^{ – 34}} \times \,\,3\,\, \times \,\,{{10}^8}}}{{1.6\,\, \times \,\,{{10}^{ – 19}} \times \,\,{\lambda _{(A)}}\, \times \,\,{{10}^{ – 10}}}}$
$\, = \,\,\frac{{12375}}{{{\lambda _{(A)}}}}$
$\,\therefore \,\,{E_{(KeV)}}\,\, = \,\,\,\frac{{12.37}}{{{\lambda _{(A)}}}}\,\, = \,\,\frac{{12.4}}{\lambda }$
$ = \,\,\frac{{hc}}{{e\lambda }}$ ($eV\,$ એકમમાં)
${E_{(eV)}}\, = \,\,\,\frac{{6.6\,\, \times \,\,{{10}^{ – 34}} \times \,\,3\,\, \times \,\,{{10}^8}}}{{1.6\,\, \times \,\,{{10}^{ – 19}} \times \,\,{\lambda _{(A)}}\, \times \,\,{{10}^{ – 10}}}}$
$\, = \,\,\frac{{12375}}{{{\lambda _{(A)}}}}$
$\,\therefore \,\,{E_{(KeV)}}\,\, = \,\,\,\frac{{12.37}}{{{\lambda _{(A)}}}}\,\, = \,\,\frac{{12.4}}{\lambda }$
Standard 12
Physics
