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The power radiated by a black body is P and it radiates maximum energy at wavelength, \lambda _{o}. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \frac{3}{4}\lambda _{o}, the power

becomes nP. The value of n

 

  • Option 1)

    \frac{256}{81}

  • Option 2)

    \frac{4}{3}

  • Option 3)

    \frac{3}{4}

  • Option 4)

    \frac{81}{256}

 

Answers (1)

best_answer

As we have learned

Radiant Power (P) -

P=frac{Q}{t}=Aepsilon sigma 	heta^{4}

-

 

 \lambda T= costant

\lambda_{0} T_{1}=3/4 \lambda _{0}T_{2}\Rightarrow T_{2}=4/3T_{1}

p\propto T_{1}^4{}

np\propto (4/3 T_{1})^{4}= (256/81)T_{1}^4

np/p = (256/81)  or n= 256/81

 

 


Option 1)

\frac{256}{81}

This is correct

Option 2)

\frac{4}{3}

This is incorrect

Option 3)

\frac{3}{4}

This is incorrect

Option 4)

\frac{81}{256}

This is incorrect

Posted by

prateek

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