The radiation corresponding to `3 to 2` transition of hydrogen atom falls on a metal surface to produce photoelectrons.These electrons are made to enter a magnetic field of `3xx10^(-4)`T.If the radius of the largest circular path followed by these electrons is 10.0 mm the work function of the metal is close to:

Options

A. 1.1 eV
B. 0.8 eV
C. 1.6 eV
D. 1.8 eV

Answer

A

Explanation

The energy in n = 3 orbit `E_(3)=-(13.6)/(9)=-1.51eV`
The energy in n=2 orbit `E_(2)=-(13.6)/(4)=-3.4eV`
The energy in n=1 orbit `E_(1)=-13.6eV`
The energy of emitted photons during the transition `3to2` of electron,
`hf=E_(3)-E_(2)`
`=-1.51-(-3.4)=3.4-1.51`
A moving electron moves in a circle in a magnetic field,
`:.(mv^(2))/(r)=Bqv`
`:.mv=Bqr`
`:.p=Bqr` and
`K_(max)=(p^(2))/(2m)=((Bqr)^(2))/(2m)` in joule
`:.K_(max)=((Bqr)^(2))/(2mxx1.6xx10^(-19))eV`
`=((3xx10^(-4)xx1.6xx10^(-19)xx10^(-2))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19))`
`=(2304)/(29.12)xx10^(-2)`
`=0.791eV" ".......(2)`
Now, from photoelectric equation of Einstein,
`K_(max)=hf-phi`
`:.phi=hf-K_(max)`
`=1.89-0.791" "` [From equ. (1) and (2) ]
`=1.099 eV`
`:.phi~~1.1eV`