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Physics Question on work, energy and power

An XX-ray tube produces a continuous spectrum of radiation with its shortest wavelength of 45×10−2?45 \times 10^{-2}?. The maximum energy of a photon in the radiation in eVeV is (h=6.62×10−34 J−s,c=3×108 m/s)(h = 6.62 \times 10^{-34} \, J-s, c = 3 \times 10^8 \, m/s)

A

27500

B

22500

C

17500

D

12500

Answer

27500

Explanation

Solution

E=hcλE = \frac{hc}{\lambda} =6.62×10−34×3×10845×10−12= \frac{6.62 \times10^{-34} \times 3 \times 10^{8}}{45\times 10^{-12}} =0.44×10−141.6×10−19= \frac{0.44\times 10^{-14}}{1.6\times 10^{-19}} =0.275×105eV= 0.275 \times 10^{5} eV =0.275×105eV= 0.275 \times 10^{5} eV