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The circumference of the second Bohr orb...

The circumference of the second Bohr orbit of electron in hydrogen atom is `600nm` . The potential difference that must be applied between the plates so that the electron have the de Broglie wavelength corresponding in this circumference is

Text Solution

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`n lamda=2pir`
`implies2xxlamda=600`
`lamda=300nm`
Let stopping potential is `V_(0)`
`eV_(0)=1/2 mv^(2)`…………1
`lamda=h/(mv)`
`:.V=h/(mlamda)`………..2
From equation 1 and 2
`eV_(0)=1/2m(h/(lamdam))^(2)`
`V_(0)=(h^(2))/(2mlamda^(2)e)`
`=((6.626xx10^(-34))^(2))/(2xx(9.1xx1xx0^(-31))xx(300xx10^(-9))^(2)xx1.6xx10^(-19)`
`=1.675xx10^(-5)V`
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