A beam of 450 nm light is incident on a metal having work function 2 eV and placed in a magnetic field B . If the most energetic electrons emitted are bent into circular are of radius 0.2 m , find B .

Question

A beam of 450 nm light is incident on a metal having work function 2 eV and placed in a magnetic field B . If the most energetic electrons emitted are bent into circular are of radius 0.2 m , find B . (A)  2.36 × 10-4 T  (B)  1.46 × 10-5 T  (C)  6.9 × 10-5 T  (D)  9.2 × 10-6 T

Tardigrade Admin · Accepted Answer

(1/2) mv2 + φ = E = h (c/λ) (1/2)mv2 + 2 × 1.6 × 10-19 = 6.6 × 10-34 × (3× 108/450 × 10-9) mv2 = (6.6 × 10-34 × 3 × 108 × 2/450 × 10-9) - 4 × 1.6 × 10-19 (1/2) × 1.9 × 10-31 v2 = 8.8 × 10-19 - 4 × 1.6 × 10-19 v2 = (4.8 × 10-19/1.9× 10-31) v = 1.58 × 106

Q. A beam of $450 nm$ light is incident on a metal having work function $2 e V$ and placed in a magnetic field $B$ . If the most energetic electrons emitted are bent into circular are of radius $0.2 m$ , find $B$ .

$2.36 \times 1 0^{- 4} T$

$1.46 \times 1 0^{- 5} T$

$6.9 \times 1 0^{- 5} T$

$9.2 \times 1 0^{- 6} T$

Q. A beam of 450nm light is incident on a metal having work function 2eV and placed in a magnetic field B . If the most energetic electrons emitted are bent into circular are of radius 0.2m , find B .

2.36×10−4T

1.46×10−5T

6.9×10−5T

9.2×10−6T

21​mv2+ϕ=E =hλc​ 21​mv2+2×1.6×10−19=6.6×10−34 ×450×10−93×108​ mv2=450×10−96.6×10−34×3×108×2​ −4×1.6×10−19 21​×1.9×10−31v2=8.8×10−19 −4×1.6×10−19 v2=1.9×10−314.8×10−19​ v=1.58×106

Q. A beam of $450 nm$ light is incident on a metal having work function $2 e V$ and placed in a magnetic field $B$ . If the most energetic electrons emitted are bent into circular are of radius $0.2 m$ , find $B$ .

$2.36 \times 1 0^{- 4} T$

$1.46 \times 1 0^{- 5} T$

$6.9 \times 1 0^{- 5} T$

$9.2 \times 1 0^{- 6} T$