If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is

  • A
    $\frac{\lambda}{2}$
  • B
    $\frac{\lambda}{3}$
  • C
    $\frac{\lambda}{4}$
  • D
    $\frac{\lambda}{9}$

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