The circumference of the second bohr orbit of electron in the h atom is 600nm. calculate the potential diff to which the electron has be subjected so that the electron stops .the electron had the de brogile wavelength , corresponding to the circumference

The circumference of the second bohr orbit of electron in the h atom i

1.	The circumference of the second bohr orbit of electron in the h atom is 600nm. calculate the potential diff to which the electron has be subjected so that the electron stops .the electron had the de brogile wavelength , corresponding to the circumference
Answer» We KNOW that 2 pi r = n LAMBDA 600 × 10 ^ -6 = 2 lambda Lambda = 300 * 10^-6 Potential V = h^2/ 2mq(lambda)^2 P = (6.6310^-34)^2/ 2 9.1×10^-31 × 1.6×10^-19 ×(600*10^-9)^2 P= 1.675×10^-5 V

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The circumference of the second bohr orbit of electron in the h atom is 600nm. calculate the potential diff to which the electron has be subjected so that the electron stops .the electron had the de brogile wavelength , corresponding to the circumference

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