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Hi, I think youve misstyped the exponent for the activity..i got 1.66*10^8 Also part d) asks for the total energy of the emiited alpha particles, the energy of one alpha is given in the question as 5.31MeV. so it think this energy needs to multiplied by the number of alspha particles emitted in...

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Im sorry the online test changes the values after you repeat the question Ive changed it: The distance between the first and fifth minima of a single-slit diffraction pattern is 0.450 mm with the screen 43.0 cm away from the slit, when light of wavelength 560 nm is used. Find the slit width.yes...

Ive been marked wrong but not sure why: The distance between the first and fifth minima of a single-slit diffraction pattern is 0.400 mm with the screen 41.0 cm away from the slit, when light of wavelength 570 nm is used. Find the slit width. y=m(theata)D/a m-order = 5 D-distance...