The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5 × 1 0 − 4 T . Assume tha

Question

The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5 × 1 0 − 4 T . Assume that the radius of the largest circular path followed by these electrons is 7 mm , the work function of the metal is: (Mass of electron = 9.1 × 1 0 − 31 k g )

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3 → 2 ⇒ 1.89 e V 5 × 1 0 − 4 T r = 7 mm r = qB m v ​ ⇒ m v = q r B ⇒ E = 2 m P 2 ​ = 2 m ( qRB ) 2 ​ = 2 × 9.1 × 1 0 − 31 J o u l e ( 1.6 × 1 0 − 19 × 7 × 1 0 − 3 × 5 × 1 0 − 4 ) 2 ​ = 18.2 × 1 0 − 31 × 1.6 × 1 0 − 19 3136 × 1 0 − 52 ​ e V = 1.077 e V We know work function = energy incident − ( K E ) electron ​ ϕ = 1.89 − 1.077 = 0.813 e V

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