A black body emits radiations of maximum intensity at 5000 A 0 when its temperature is 1227- C - If its temperature is increased by 1000 - C then the maximum intensity of emitted radiation will be at -

The correct option is B 3000 A ^ 0 [ λ _ 1 T_ 1 =λ _ 2 T_ 2; ⇒( 5000 ) ( 1500 ) =λ _ 2 ( 2500 ) #160 ...

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Standard X

Biology

Health Hazards from Radiations

A black body ...

Question

A black body emits radiations of maximum intensity at $5000 A^{0}$ when its temperature is $1227^{\circ} C .$ If its temperature is increased by $1000^{\circ} C$ then the maximum intensity of emitted radiation will be at :

2754.8 A^{0}

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3000 A^{0}

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3500 A^{0}

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4000 A^{0}

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Solution

The correct option is B $3000 A^{0}$
$\begin{matrix} λ_{1} T_{1} = λ_{2} T_{2} \Rightarrow (5000) (1500) = λ_{2} (2500) \Rightarrow λ_{2} = \frac{5000 \times 15}{25} = 3000 A^{0} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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A black body emits radiations of maximum intensity at 5000A0 when its temperature is 1227∘C. If its temperature is increased by 1000∘C then the maximum intensity of emitted radiation will be at :

The correct option is B 3000A0λ1T1=λ2T2⇒(5000)(1500)=λ2(2500)⇒λ2=5000×1525=3000A0Hence,optionBiscorrectanswer.

A black body emits radiations of maximum intensity at $5000 A^{0}$ when its temperature is $1227^{\circ} C .$ If its temperature is increased by $1000^{\circ} C$ then the maximum intensity of emitted radiation will be at :

The correct option is B $3000 A^{0}$
$\begin{matrix} λ_{1} T_{1} = λ_{2} T_{2} \Rightarrow (5000) (1500) = λ_{2} (2500) \Rightarrow λ_{2} = \frac{5000 \times 15}{25} = 3000 A^{0} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$