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Chapter 38: Q. 47 (page 1116)

Electrons, all with the same speed, pass through a tiny 15-nm-wide slit and create a diffraction pattern on a detector 50 mm behind the slit. What is the electron鈥檚 kinetic energy, in eV, if the central maximum has a width of 3.3 mm?

Short Answer

Expert verified

The electron鈥檚 kinetic energy isEk=2507eV

Step by step solution

01

Given information

The diffraction pattern on a detector 50 mm behind the slit .

02

Making diagram of diffraction pattern 

03

Calculations

Starting with a diffraction expression on one slit, we can solve the problem.

dsin鈦�胃n=苍位dsin鈦�胃=位(n=1central maximum)tan鈦�胃=螖测2Lfrom picture胃=arctan鈦�螖测2L(express胃)位=dsin鈦�arctan鈦�螖测2L(substitute胃)Ek=hc位(kinetic energy)Ek=hcdsin鈦�arctan鈦�螖测2L(substitute位)Ek=6.63脳10鈭�34脳3脳10815鈰�10鈭�9鈰�sin鈦�arctan鈦�3.3脳10鈭�32鈰�50脳10鈭�3(substitute)Ek=2507eV

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