Question

# The de-Broglie wavelength of a neutron at $${ 27 }^{ o }C$$ is $$\lambda$$. What will be its wavelength at $${ 927 }^{ o }C$$?

A
λ/2
B
λ/3
C
λ/4
D
λ/9

Solution

## The correct option is A $$\lambda /2$$$$\quad { \lambda }_{ neutron }\propto \cfrac { 1 }{ \sqrt { T } } \Rightarrow \cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\sqrt { \cfrac { { T }_{ 2 } }{ { T }_{ 1 } } }$$$$\Rightarrow \cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\sqrt { \cfrac { (273+927) }{ (273+27) } } =2\Rightarrow { \lambda }_{ 2 }=\cfrac { \lambda }{ 2 }$$Physics

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