An electron is at a distance of 10 m from a charge of 10c.Its total energy is `15.6xx10^(-10)` J.Its de-Broglie wavelength at this point is…….. .`(h=6.625xx10^(-34))`J.s ,`m_(e)=9.1xx10^(-31)kg, K=9xx10^(9)SI`

Options

A. 9.87 Å
B. 9.87 fermi
C. 8.97 Å
D. 8.97 fermi

Answer

D

Explanation

Potential energy of an electron,U=`-(kq(e))/(r )`
`therefore U=(9xx10^(9)xx10xx1.6xx10^(-19))/(10)`
`therefore U=-14.4xx10^(-10)J`
Now total energy E=Kinetic energy K+ Potential energy U
`therefore K=E-U`
`=15.6xx10^(-10)+14.4xx10^(-10)`
`therefore K=30xx10^(-10)J`
But K=`(p^(2))/(2m_(2))`
`therefore p=sqrt(2Km_(e))`
`lambda=(h)/(sqrt(2Km_(e)))=(6.625xx10^(-34))/(sqrt(2xx30xx10^(-10)xx9.1xx10^(-31)))`
`therefor lambda=8.97xx10^(-15)m`