If the KE of electron is `2.5 xx 10^(-24)` J , then calculate its de-Broglie wavelength.

Explanation

Kinetic energy `=(1)/(2) mv^(2)`
` V=((2KE)/( m ))^(1//2)`
since , mass of electron `= 9.1 xx 10 ^(-31)Kg`
then substituting the values
` V=((2xx2.5xx10 ^(-24) J)/( 9.1 xx 10 ^(-31) Kg ))^(1//2). (1J = Kg m^(2) s^(-2))`
` v= 2.34 xx 10 ^(3) ms^(-1)`
Wavelength `(lamda ) = (h )/(mv )`
` =( 6.626 xx 10 ^(-34)JS)/( (9.1 xx 10 ^(-31)kg) ( 2.34 xx10 ^(3) ms^(-1)))`
` = 311.1 xx 10 ^(-9) m, (1nm = 10 ^(-9) m )`
` lamda = 311.1 nm `