Question 281:
easyThe ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
(2018)
The relationship between kinetic energy $K$, potential energy $U$, and total energy $E$ for an electron in a Bohr orbit is $K = -E$ and $U = 2E$. Therefore, the ratio of kinetic energy to total energy is $1 : -1$.
Question 282:
easyThe ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
(2017-Delhi)
The last line of a series corresponds to $n_2 = \infty$. For the Balmer series, $1/\lambda_B = R(\frac{1}{2^2} - 0) \Rightarrow \lambda_B = \frac{4}{R}$. For the Lyman series, $1/\lambda_L = R(\frac{1}{1^2} - 0) \Rightarrow \lambda_L = \frac{1}{R}$. The ratio is $\lambda_B / \lambda_L = 4$.
Question 283:
easyIf an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:
(2016 – II)
For the first transition, $1/\lambda = R(\frac{1}{2^2} - \frac{1}{3^2}) = \frac{5R}{36}$. For the second transition, $1/\lambda' = R(\frac{1}{3^2} - \frac{1}{4^2}) = \frac{7R}{144}$. Dividing the two gives $\lambda' = \lambda \times \frac{144/7}{36/5} = \frac{20}{7}\lambda$.
Question 284:
easyGiven the value of Rydberg constant is $10^7 m^{-1}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be:
(2016 – I)
The wave number $\bar{\nu}$ is $1/\lambda$. For the last line of the Balmer series, $n_1 = 2$ and $n_2 = \infty$. Thus, $\bar{\nu} = R(\frac{1}{2^2} - 0) = \frac{R}{4} = \frac{10^7}{4} = 0.25 \times 10^7 m^{-1}$.
Question 285:
easyThe total energy of an electron in an atom in an orbit is -3.4 eV. Its kinetic and potential energies are, respectively:
(2019)
For an electron in an orbit, kinetic energy is equal to the negative of total energy: $K = -E = -(-3.4 eV) = 3.4 eV$. Potential energy is twice the total energy: $U = 2E = 2 \times (-3.4 eV) = -6.8 eV$.
Question 286:
easy24. The ionisation energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between: (2009)
Number of spectral lines emitted is $\frac{n(n-1)}{2} = 6 \Rightarrow n = 4$. Maximum wavelength corresponds to the minimum energy difference. For transitions among levels up to $n=4$, the transition $4 \rightarrow 3$ has the minimum energy difference and thus the maximum wavelength.
Question 287:
easyThe ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is:
(2008)
Excitation energy is the energy required to excite the electron from the ground state ($n=1$) to a particular state. The energy of the first excited state ($n=2$) is $-13.6/4 = -3.4 eV$. The excitation energy is $-3.4 - (-13.6) = 10.2 eV$.
Question 288:
easyThe total energy of electron in the ground state of hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is:
(2007)
The total energy in the first excited state ($n=2$) is $E_2 = -13.6/2^2 = -3.4 eV$. Since kinetic energy $K = -E_n$, the kinetic energy in the first excited state is $3.4 eV$.
Question 289:
easyIonisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be:
(2006)
Initial energy is $-13.6 eV$. After absorbing $12.1 eV$, the final energy is $-13.6 + 12.1 = -1.5 eV$. This corresponds to the $n=3$ state (since $-13.6/3^2 \approx -1.51 eV$). Number of spectral lines emitted upon returning to ground state is $\frac{3(3-1)}{2} = 3$.
Question 290:
easyConsider 3rd orbit of $He^+$ (Helium) using non relativistic approach the speed of electron in this orbit will be (given $K = 9 \times 10^9$ constant $Z = 2$ and h (Planck’s constant) = $6.6 \times 10^{-34} Js$):
(2015)
Speed of electron in nth orbit is $v_n = 2.18 \times 10^6 \frac{Z}{n} m/s$. For $He^+$, $Z=2$ and $n=3$. So, $v_n = 2.18 \times 10^6 \times \frac{2}{3} = 1.453 \times 10^6 m/s \approx 1.46 \times 10^6 m/s$.