Question

The photoelectric work function of a metal is 3eV .Find the maximum kinetic energy and maximum speed photoelectrons when radiation of wavelength 4000A is incident on the metal surface.
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Answer 1

Answer:

Find the maximum K.E. and the maximum speed of photoelectrons emitted when radiation of wavelength 4000 Å is incident on the metal surface. [Given: Mass of electron = 9.1 x 10-lkg] [Mar 00] Ans: 1.725 x 10-20 J, 1.947 x 109 m/s.

Answer 2

Answer:

maximum kinetic energy 1.725 × 10⁻²⁰ J is and maximum speed of photoelectrons is 1.947 × 10⁵ m/s

Explanation:

Here given,

photoelectric work function of a metal, Q = 3eV = 3 × 1.6 × 10⁻¹⁹

radiation of wavelength incident on the metal surface, λ =  4000A

speed of light, c = 3 × 10⁸ m/s

plank's constant, h = 6.6 × 10⁻³⁴

maximum kinetic energy,K.E. = (hc/λ) - Q

= [(6.6 × 10⁻³⁴ × 3 × 10⁸)/4000] - 3 × 1.6 × 10⁻¹⁹

= 1.725 × 10⁻²⁰ J

Also we know that,

1/2mv² = K.E.

v = √(2K.E. / m)

= √[(1.725 × 10⁻²⁰ × 2) / 9.1 × 10⁻³¹]

= 1.947 × 10⁵ m/s

So, the maximum kinetic energy 1.725 × 10⁻²⁰ J is and maximum speed of photoelectrons is 1.947 × 10⁵ m/s