Question

When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is ______ A°. (Round off to the Nearest Integer)

[Use: `sqrt3` = 1.73, h = 6.63 × 10⁻³⁴ Js, m_e = 9.1 × 10⁻³¹ kg, c = 3.0 × 10⁸ ms⁻¹, 1eV = 1.6 × 10⁻¹⁹ J]

पर्याय

• 6

• 7

• 8

• 9

Answer 1

When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is 9 A°.

Explanation:

Threshold energy (W₀) = 3 eV

= 3 × 1.6 × 10⁻¹⁹ J

Wavelength (λ) = 248 nm

= 248 × 10⁻⁹ m

E = W₀ + KE

KE = E − W₀

= `"hc"/λ - "W"_0`

= `((6.6 xx 10^-34 xx 3 xx 10^8)/(248 xx 10^-9)) - 3 xx 1.6 xx 10^-19`

= 3.2 × 10⁻¹⁹ J

P = `sqrt(2"mKE")`

= `sqrt(2 xx 9.1 xx 10^-31 xx 3.2 xx 10^-19)`

= 7.633 × 10⁻²⁵ 

λ = `"h"/"P"`

= `6.6 xx 10^-34/7.63 xx 10^-25`

= 8.7 × 10⁻¹⁰

= 8.7 A°

≈ 9