JEE MAIN - Physics (2005 - No. 25)
If the kinetic energy of a free electron doubles, it's deBroglie wavelength changes by the factor
$$2$$
$${1 \over 2}$$
$${\sqrt 2 }$$
$${1 \over {\sqrt 2 }}$$
توضيح
de-Broglie wavelength,
$$\lambda = {h \over p} = {h \over {\sqrt {2.m,\left( {K.E} \right)} }}$$
$$\therefore$$ $$\lambda \propto {1 \over {\sqrt {K.E} }}$$
If $$K.E$$ is doubled, wavelength becomes $${\lambda \over {\sqrt 2 }}$$
$$\lambda = {h \over p} = {h \over {\sqrt {2.m,\left( {K.E} \right)} }}$$
$$\therefore$$ $$\lambda \propto {1 \over {\sqrt {K.E} }}$$
If $$K.E$$ is doubled, wavelength becomes $${\lambda \over {\sqrt 2 }}$$