According to Bohr-s Theory of hydrogen atom- the product of the binding energy of the electron in the nth orbit and its radius in the nth orbitIs proportional to n-2Is inversely proportional to n-3Has a constant value of 10-2space eV-mathring-A-Has a constant value of 7-2space eV-mathring-A-

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Toppr Admin · Accepted Answer

From the formulas attached-The productR-n-xD7-E-n-0-529n2Z-xD7-13-6Z2n2eV-x2212-AR-n-xD7-E-n-0-529-xD7-13-6ZeV-x2212-AR-n-xD7-E-n-7-2eV-x2212-A

According to Bohr's Theory of hydrogen atom, the product of the binding energy of the electron in the $n t h$ orbit and its radius in the $n t h$ orbit
Is proportional to $n^{2}$
Is inversely proportional to $n^{3}$
Has a constant value of $10.2 e V - ˚ A$
Has a constant value of $7.2 e V - ˚ A$

From the formulas attached.
The product
$R (n) \times E (n) = 0.529 \frac{n^{2}}{Z} \times 13.6 \frac{Z^{2}}{n^{2}} e V . - A$
$R (n) \times E (n) = 0.529 \times 13.6 Z e V . - A$
$R (n) \times E (n) = 7.2 e V - A$

According to Bohr's Theory of hydrogen atom, the product of the binding energy of the electron in the nth orbit and its radius in the nth orbitIs proportional to n2Is inversely proportional to n3Has a constant value of 10.2 eV−˚AHas a constant value of 7.2 eV−˚A

From the formulas attached.The productR(n)×E(n)=0.529n2Z×13.6Z2n2eV.−AR(n)×E(n)=0.529×13.6ZeV.−AR(n)×E(n)=7.2eV−A

According to Bohr's Theory of hydrogen atom, the product of the binding energy of the electron in the $n t h$ orbit and its radius in the $n t h$ orbit
Is proportional to $n^{2}$
Is inversely proportional to $n^{3}$
Has a constant value of $10.2 e V - ˚ A$
Has a constant value of $7.2 e V - ˚ A$

From the formulas attached.
The product
$R (n) \times E (n) = 0.529 \frac{n^{2}}{Z} \times 13.6 \frac{Z^{2}}{n^{2}} e V . - A$
$R (n) \times E (n) = 0.529 \times 13.6 Z e V . - A$
$R (n) \times E (n) = 7.2 e V - A$