Question

the maximum wavelength for photoelectric emission in tungsten is 230 nm what wavelength of light must be used in order for electrons with a maximum kinetic energy 1.7ev to be ejected?​

Answer 1

Dear Student,

◆ Answer -

λ = 175 nm

● Explaination -

# Given -

λ₀ = 230 nm = 230×10^-9 m

KE = 1.7 eV = 1.7×1.6×10^-19 J

# Solution -

Wavelength of incident light is calculated as -

KE = hc/λ - hc/λ₀

KE = hc (1/λ - 1/λ₀)

1.7×1.6×10^-19 = 6.63×10^-34 × 3×10^8 [1/λ - 1/(230×10^-9)]

1.7×1.6×10^-19 / 1.988×10^-25 = (1/λ - 4.348×10^6)

1.368×10^6 = 1/λ - 4.348×10^6

1/λ = 1.368×10^6 + 4.348×10^6

1/λ = 5.716×10^6

λ = 1.749×10^-7 m

λ = 175 nm

Therefore, wavelength of light used is 175 nm.

Thanks dear...

Answer 2

Answer:

Explanation:

pta nhio