A metal surface is illuminated by a radiation of wavelength 4500 mathringA. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately:

Question

A metal surface is illuminated by a radiation of wavelength 4500 mathringA. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately: (A) 1.36 eV (B) 1.69 eV (C) 2.78 eV (D) 2.23 eV

Tardigrade Admin · Accepted Answer

λ=4500 mathringA B =2 mT , R =2 mm R =(√2 Km / qB ) (( qBR )2/2 m )= K ((1.6 × 10-19 × 2 × 10-3 × 2 × 10-3)2/2 × 9.1 × 10-31)= K ((6.4)2/2 × 9.1) × (10-50/10-31)= K K =2.25 × 10-19 J =(2.25 × 10-19/1.6 × 10-19) eV =1.40 eV E =(12400/4500)=2.76 eV φ= E - K =(2.76-1.40) eV =1.36 eV

$1.36 e V$

$1.69 e V$

$2.78 e V$

$2.23 e V$

Q. A metal surface is illuminated by a radiation of wavelength 4500A˚. The ejected photo-electron enters a constant magnetic field of 2mT making an angle of 90∘ with the magnetic field. If it starts revolving in a circular path of radius 2mm, the work function of the metal is approximately:

1.36eV

1.69eV

2.78eV

2.23eV

λ=4500A˚ B=2mT,R=2mm R=qB2Km​​ 2m(qBR)2​=K 2×9.1×10−31(1.6×10−19×2×10−3×2×10−3)2​=K 2×9.1(6.4)2​×10−3110−50​=K K=2.25×10−19J =1.6×10−192.25×10−19​eV=1.40eV E=450012400​=2.76eV ϕ=E−K=(2.76−1.40)eV=1.36eV

$1.36 e V$

$1.69 e V$

$2.78 e V$

$2.23 e V$