Statement-1: If the potential difference applied to an electron is made $4$ times, the de-Broglie wavelength associated is halved. (Initial kinetic energy of the electron was zero).
Statement-2: On making potential difference $4$ times velocity is doubled and hence according to De-broglie hypothesis $λ$ is halved.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Statement-1 is false, statement-2 is true.

Statement-1 is true, statement-2 is false.

Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$λ_{1} = \sqrt{\frac{150}{V}} \overset{°}{A,} λ_{2} = \sqrt{\frac{150}{4 V}} \overset{°}{A} = \frac{1}{2} \sqrt{\frac{150}{V}} \overset{°}{A} = \frac{λ_{1}}{2} \frac{1}{2} m_{e} {u_{1}}^{2} = eV \frac{1}{2} m_{e} {u_{2}}^{2} = 4 eV \Rightarrow u_{1} = \sqrt{\frac{2 eV}{m_{e}}} \Rightarrow u_{2} = 2 \times \sqrt{\frac{2 eV}{m_{e}}} λ_{1} = \frac{h}{{mu}_{1}} = 2 u_{1} λ_{2} = \frac{h}{{mu}_{2}} = \frac{h}{2 {mu}_{1}} = \frac{λ_{1}}{2}$